The post Resistor Types and Color Code appeared first on Electrical A2Z.

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**Figure 1.** Group of carbon composition resistors and the fixed resistor symbol.

The chemical makeup that causes resistance is accurately controlled in the resistor manufacturing process. Resistor values can be purchased in a range of values from less than 1 ohm to over 22 mega-ohms.

The physical size and material used for resistance is rated in watts. A resistor’s wattage rating refers to the resistor’s ability to safely dissipate heat. Heat is generated by electrons flowing through the resistor. Common wattage sizes range from 1/4 watt to 25 watts. Resistors are grouped by ohms and watt sizes. See **Figure 2**.

**Figure 2.** The physical size of a resistor can vary according to wattage rating. The higher the wattage, the larger the resistor. The largest shown is 20 watts, and the smallest is 1/8 watt.

When purchasing a resistor, the desired resistance and wattage rating must be specified. For example: 1000 ohms and the watt size, 1/4 watt, 1/2 watt, or 2 watts, etc. In each watt size, the resistance value would be the same. See **Figure 3** to examine the construction of a **molded composition resistor**.

**Figure 3**. Cutaway of a carbon composition resistor. (Allen-Bradley)

**Thin Film Resistor**

Another type of small wattage, fixed value resistor is the **thin film resistor**. The thin film resistor is similar to the molded composition resistor in appearance and function. However, the thin film resistor is made by depositing a resistance material on a glass or ceramic tube. A photographic process is used to deposit this film. Leads with caps are fitted over each end of the tube to make the body of the resistor. Thin film resistors are usually color coded.

The term “film resistor” is generally used to classify very compact resistors used in micro-electronics or on very small-scale electronic circuit boards. Film resistors can also be referred to as **surface-mounted resistors (SMRs)**. The demand for smaller and smaller electronic devices, such as cell phones, created the need for small discrete components such as resistors, to be manufactured in a more compact method.

Thick film and thin film are two general classifications based on how the film resistor is manufactured. Thin film deposits resistive material on an insulated substrate. Then the undesired portion is etched away leaving the desired pattern of resistive material. See **Figure 4**.

**Thick film** deposits a special resistive paste directly on the insulated substrate by using a stencil or silk screen process. As a result, thick film is typically a thicker deposit of resistive material as compared to thin film. The advantage of thick film is the resistor can support higher currents and wattage than the thin film. The advantage of thin film is smaller components requiring less height can be made.

**Figure 4.** This cutaway drawing displays how a film resistor is constructed.

The **substrate** is made from glass, ceramic, or silicon, and it is used as an insulator base for the resistor. A layer of resistor material is deposited on the substrate in a zigzag pattern precisely engineered to produce the desired resistance value.

The **resistor materials** are made from metals or carbon mixed in a precise proprietary formula. The result is a thin film, which is only a few micrometers thick or thick film, which is 10 to 50 micrometers thick.

A **protective coating** is used to cover the resistor material deposited on the substrate. The ends of the film resistor serve as connection terminals and are made of metal such as nickel or silver, or they can be made from an alloy.

**Solder** is used to mount and connect the film resistor to the circuit board. See **Figure 5**.The process is very similar to the same process used to manufacture integrated circuits (ICs).

**Figure 5.** You can see film resistors mounted on a circuit board.

The number stenciled directly on surface-mounted resistors indicates the resistance value. When a three-digit number is used, the first two digits represent the first two digits of the resistance value and the third digit represents the number of zeros to add to the first two digits. **For example**, the 202 in **Figure 5** represents the resistance value 2000 or 2k, + or – 5% tolerance (meaning it is allowed to vary by 5%).

When **four digits are used**, a tolerance of + or – 1% exists. For four-digit numbers, the same rule applies that the last digit represents the number of zeros. **For example**, 5002 would represent 50,000 or 50 k ohms, + or – 1% tolerance. For resistance values that require a decimal, the letter “R” is used to indicate the decimal point. For example, 7R5 would represent 7.5 ohms.

Film technology is not restricted to single resistors but may also be designed to produce resistor networks on a single chip. A resistor network consists of two or more independent resistors contained in one single surface- mounted package.

**Wire Wound Resistor**

For higher current uses, resistors are **wire wound**. A thin wire is wound on a ceramic core. The wire has a specific fixed-value resistance. The entire component is insulated by a coat of vitreous (opaque) enamel. These resistors are shown in **Figure 6**.

Wire wound resistors are commonly manufactured in sizes from 5 to 200 watts. The wattage chosen depends on the heat dissipation required during operation. Metal oxide resistors are also used for high voltage and wattage requirements.

**Figure 6**. Wire wound resistors.

**Adjustable Resistor **

Another type of wire wound resistor is the adjustable resistor. Unlike the standard wire wound resistor, the adjustable resistor is not entirely covered by enamel material. Instead, a portion of one side of the wire is exposed.

An adjustable sliding tap is attached to move across the exposed surface. This allows the resistance value to be varied. Adjustable resistors may have two or more taps for providing various resistance values in the same circuit. An adjustable resistor and symbol are shown in **Figure 7**.

**Figure 7.** This adjustable resistor provides a sliding tap for voltage divider uses. On the right is the schematic symbol for an adjustable resistor.

**Potentiometers**

Most electronic equipment requires the use of variable resistance parts. A potentiometer is a very common type of variable resistor found in electronic projects. The potentiometer has a rotary knob that varies the resistance value as it is turned.

The variation in resistance is provided by a contact that is attached to a ring of resistive material inside the device. This device is similar to the wire wound resistor.

Many potentiometers are constructed with thin wire inside as the source of resistance. Various styles of potentiometers are illustrated in **Figure 7a**. Also shown is the potentiometer’s schematic symbol.

**Figure 7a.** Potentiometers are used in electronic circuitry for fixed and variable resistance. Notice that the schematic symbol for the potentiometer is the same as for the adjustable resistor.

**Thermistors**

A special type of resistor is called a thermistor. In comparison to other types of resistors, the thermistor is unusual due to its ability to change resistance value rapidly as its temperature changes. It is commonly used to prevent high inrush currents in electrical circuits.

**An example** of a thermistor use can be seen in a **blow dryer**. A common blow dryer has heating elements composed of tungsten wire. The tungsten wire has a very low resistance value when cold, and a high resistance value when red hot. The thermistor is placed in series with the heating elements to prevent a high current value when the dryer is first turned on. As the blow dryer heats up, the resistance value goes down. The result is a fairly consistent current value as the dryer’s heating element changes from low resistance (cold) to high resistance (hot).

Early blow dryer models caused a dimming and flickering of lights and other electronic equipment in the home because of inconsistent current draw. The thermistor eliminates this problem.

Larger resistors are usually marked with their numerical resistance value printed directly on the side of the resistor. However, this type of labeling is not always practical, especially on small resistors. The resistor color code system was developed for this purpose.

The color code marking system has been adopted by the Electronics Industries Association (EIA) and the United States Armed Forces. This system of color coding is recognized throughout the world. Refer to **Figure 8**. Note how the color codes are printed, or banded, around the entire body of the resistor. This method of coding permits the value of the resistor to be read regardless of the mounting position. To see how to read the color coded bands refer to **Figure 9**.

F**igure 8. **Color coded bands encircle the resistor.

**Figure 9.** Standard color code for resistors.

Resistors commonly have three or four (and sometimes five) bands. Each band has a unique meaning.

- The
**first band**represents the value of the first digit of the resistance value. - The
**second band**represents the second digit of the resistance value. - The
**third band**is called the multiplier. The multiplier gives the factor of ten that the first two digits should be multiplied by. - The
**fourth band**represents the tolerance of the resistor.

Resistor tolerance is a reflection of the precision of resistor’s value. If a 20 ohm resistor has a 10% tolerance, the resistor’s value can vary by ± 2 ohms. In this case the resistor can have a true resistance value of 18 to 22 ohms. A fifth band is sometimes used to indicate resistor reliability or expected failure rate.

See **Figure 10**. In the first sketch, the **first bar** is red and the second bar is violet. Checking the color code chart shows the first two digits in the resistor’s value to be 2 and 7 (or 27). This number is then multiplied by the **third band**. The third band is brown (× 10). This indicates that the value 27 is multiplied by 10, for a final value of 270 ohms. This is what the resistor value would be if the resistor was perfect. **However**, the **fourth band** is silver. This indicates that this resistor has a tolerance of 10 percent. **In this example**, the tolerance is a ±27 ohms (270 × 0.1 = 27). Thus the value of this resistor is somewhere between 243 ohms and 297 ohms.

Work out the stated value, maximum value, and minimum value for the other two sketches in **Figure 10** on your own.

**Figure 10**. Examples of standard color-coded resistors.

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]]>The post Electrical Circuit Devices appeared first on Electrical A2Z.

]]>**Switches**

Switches are installed in circuits to control the flow of electrons through the circuit. They can be categorized by their actuator and electrical switching path. The actuator is the mechanical device that causes the circuit to open and close.

Various types of switches are illustrated in **Figure 1**. The schematic symbol associated with each type of switch is also included.

**Figure 1.** Typical switch types. Notice how many have the same electrical symbol but different actuators.

Some of the most common actuators are the slide, toggle, rotary, and **push button**. As you can see, the name of the switch indicates the type of actuator used to turn the circuit on and off.

The electrical circuit inside the switch is described in terms of poles and throws. The simplest type of switch is the **single-pole single-throw switch**, which is abbreviated SPST. The term single-pole means that the switch provides one path for the electron flow and that it can be turned on or off. The term single-throw means that the switch controls only one circuit.

A **single-pole double-throw switch (SPDT)** has one common connection point and can complete a circuit path to two different circuits. However, only one circuit can be completed at a time. There are many possibilities and combinations for a switch of this type.

A useful application of the single-pole double-throw switch is its ability to control a load, such as a lamp, from two different locations. In residential wiring systems, the single-pole double-throw switch is referred to as a **three-way switch**.

In **Figure 2**, two single-pole double-throw switches are connected to a lamp. Either switch is capable of turning the lamp on or off.

**Figure 2.** Two single-pole double-throw switches (SPDT) can be used to control a lamp from two different locations.

A double-pole double-throw switch (DPDT) has two common connection points and can provide two circuit paths simultaneously. The DPDT switch is like having two SPST switches connected in parallel.

Switches are rated for **ampacity and voltage**.

The

ampacityrating of a switch is an indication of how much current it can safely handle.The

voltage ratingis the maximum voltage for which a switch is designed.

Exceeding the maximum voltage rating will cause the electrical-mechanical circuitry inside the switch to fail. **For example**, if a toggle switch is rated as one amp and 24 volts, a current in excess of one amp will burn out the switching circuitry inside the switch. If the 24-volt switch circuit sufficiently to stop the flow of electrons. This action will result in a dangerous situation that can melt the switch’s insulation and short circuit the switch.

**Connectors**

There are many types of connectors used with electrical conductors. The type of connection used varies according to the type and size of the conductor, the purpose served by the connection, and the type of device to be connected.

Look at **Figure 3**. You will see many common types of connectors. One general classification is solder less connectors. A solder less connector does not require the use of solder to make the connection. These connectors generally require a crimping tool. The crimping tool squeezes the connector to the conductor. **Figure 3** shows common wire crimps on terminals and splices.

**Figure 3**. Some wire connectors are made to be crimped on the end of stranded conductors. After the connector is crimped on the wire, the wire can be easily secured under a termination block screw.

Some types of connectors use screws and bolts to form the mechanical connection to conductors. These connectors are used primarily for larger conductors. See **Figure 4**.

**Figure 4.** Two types of solder less connectors are the wire nut connector and the split-bolt connector. The wire nut is used extensively in residential and commercial wiring. The split-bolt connector is used mainly on large diameter conductors.

Common circuit protection devices are fuses and circuit breakers.

**Fuse**

Fuses, such as those shown in **Figure 5**, are constructed from small, fine wire. This wire is engineered to burn if certain amperages are exceeded. Fuses are sized by their voltage and current capacity, primarily current.

**Figure 5.** A typical fuse and the schematic symbol that represents the fuse.

**For example, **a three-amp fuse is designed to burn and open the circuit when the current exceeds three amps. A load that draws three amps or greater will generate sufficient heat in the fuse to melt the fuse link inside the glass tube. The **time required** to melt the fuse link is inversely proportional to the amount of overload. This means that the higher the overload current, the faster the melting action occurs. When a fuse melts, it must be replaced.

**Circuit breaker**

A **circuit breaker**, sometimes called a reset, is another device used to protect a circuit from overload and short circuit conditions. See **Figure 6**.

The main advantage of a circuit breaker over the fuse link is that the circuit breaker need not be replaced after tripping. It can be reset by moving the handle to the off position and then returning it to the on position.

Some circuit breakers have an actuator similar to a push button switch. These breakers are pushed in to reset after tripping. Circuit breakers and resets require a waiting period to allow the internal trip mechanism to cool down.

Most homes today use circuit breakers as the safety device to prevent overloads. Overloads could result in house fires.

**Figure 6. **A typical circuit breaker.

Circuit breakers are produced with two different tripping methods.

**One method** uses bimetallic strips. A bimetallic strip is a metal strip made of two different types of metal. Different metals expand at different rates. Heat generated from the overload condition causes the bimetal trip to expand. The different metals expand at different rates. This causes the breaker’s trip mechanism to bend and break contact. Some trip mechanisms are adjustable to allow for a more precise trip current.

A **second tripping mechanism** uses magnetism to operate. The circuit current runs through a coil. As the current increases through the coil, the amount of magnet- ism in the coil increases. When a predetermined point is reached, the tripping mechanism operates and opens the circuit.

The magnetic circuit breaker is much faster and more accurate than the bimetallic circuit breaker. The magnetic circuit breaker, however, is more expensive than the bimetallic type.

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]]>The post Low Pass and High Pass Filter Bode Plot appeared first on Electrical A2Z.

]]>\[\begin{matrix}{{\left| \frac{{{A}_{0}}}{{{A}_{i}}} \right|}_{dB}}=20{{\log }_{10}}\left| \frac{{{A}_{0}}}{{{A}_{i}}} \right|=20{{\log }_{10}}\left| \frac{{{A}_{0}}}{{{A}_{i}}} \right| & {} & (1) \\\end{matrix}\]

While logarithmic plots may at first seem a daunting complication, they have two significant advantages:

1. The product of terms in a frequency response function becomes a sum of terms because log(*ab/c*) = log(*a*) + log(*b*) − log(*c*). The advantage here is that Bode (logarithmic) plots can be constructed from the sum of individual plots of individual terms. Moreover, there are only four distinct types of terms present in any frequency response function:

a. A constant *K*.

b. Poles or zeros “at the origin”(*jω*).

c. Simple poles or zeros (1 + *jωτ*) or (1 + *jω/ω _{o}*).

d. Quadratic poles or zeros [1+ *jωτ* +(*jω/ω _{n}*)

2. The individual Bode plots of these four distinct terms are all well approximated by linear segments, which are readily summed to form the overall Bode plot of more complicated frequency response functions.

Consider the RC low-pass filter. The frequency response function is:

\[\frac{{{V}_{0}}}{{{V}_{i}}}(j\omega )=\frac{1}{1+j\omega /{{\omega }_{0}}}=\frac{1}{\sqrt{1+{{(\omega /{{\omega }_{0}})}^{2}}}}\angle -{{\tan }^{-1}}\left( \frac{\omega }{{{\omega }_{0}}} \right)\begin{matrix}{} & (2) \\\end{matrix}\]

where the circuit time constant is τ = *RC* = 1/*ω _{0}* and

**Figure 1** shows the Bode magnitude and phase plots for the filter.

**Figure 1** Bode plots for a low-pass *RC* filter; the frequency variable is normalized to *ω*/*ω*0. (a) Magnitude response; (b) phase angle response

The normalized frequency on the horizontal axis is *ωτ*. The magnitude plot is obtained from the logarithmic form of the absolute value of the frequency response function.

\[{{\left| \frac{{{V}_{0}}}{{{V}_{i}}} \right|}_{dB}}=20{{\log }_{10}}\frac{\left| K \right|}{\left| 1+j\omega \tau \right|}=20{{\log }_{10}}\frac{\left| K \right|}{\left| 1+j\omega /{{\omega }_{0}} \right|}\begin{matrix}{} & (3) \\\end{matrix}\]

When *ω ≪ ω _{0}*, the imaginary part of the simple pole is much smaller than its real part, such that |1 +

\[\begin{matrix}{{\left| \frac{{{V}_{0}}}{{{V}_{i}}} \right|}_{dB}}\approx 20{{\log }_{10}}K-20{{\log }_{10}}1=20{{\log }_{10}}K & (dB) & (4) \\\end{matrix}\]

Thus, at very low frequencies (*ω ≪ ω _{0}*),

When *ω ≫ ω*_{0}, the imaginary part of the simple pole is much larger than its real part, such that |1 + *jω/ω _{0}| ≈ | jω/ω_{0}*| = (

\[\begin{matrix}{{\left| \frac{{{V}_{0}}}{{{V}_{i}}} \right|}_{dB}}\approx 20{{\log }_{10}}K-20{{\log }_{10}}\frac{\omega }{{{\omega }_{0}}} & {} & {} \\{} & {} & (5) \\\approx 20{{\log }_{10}}K-20{{\log }_{10}}\omega +20{{\log }_{10}}{{\omega }_{0}} & {} & {} \\\end{matrix}\]

Thus, at very high frequencies (*ω ≫ ω _{0}*),

Finally, when *ω = ω _{0}*, the real and imaginary parts of the simple pole are equal, such that |1 +

\[20{{\log }_{10}}\frac{\left| K \right|}{\left| 1+j\omega /{{\omega }_{0}} \right|}=20{{\log }_{10}}K-20\log \sqrt{2}=20{{\log }_{10}}K-3dB\begin{matrix}{} & (6) \\\end{matrix}\]

Thus, the Bode magnitude plot of a **first-order low-pass filter** is approximated by two straight lines intersecting at *ω _{0}*.

The phase angle of the frequency response function $\angle \left( \frac{{{V}_{o}}}{{{V}_{i}}} \right)=-{{\tan }^{-1}}\left( \frac{\omega }{{{\omega }_{o}}} \right)$ has the following properties:

As a first approximation, the phase angle can be represented by three straight lines:

- For ω < 0.1ω
_{o}, ∠ (V_{o}/V_{i}) ≈ 0. - For 0.1 ω
_{o}and 10ω_{o}, ∠ (V_{o}/V_{i}) ≈ – (π/4) log (10ω/ω_{o}). - For ω > 10ω
_{o}, ∠ (V_{o}/V_{i}) ≈ –pi/2.

These straight-line approximations are illustrated in **Figure 1(b).**

**Table 1** lists the differences between the actual and approximate Bode magnitude and phase plots. Note that the maximum difference in magnitude is 3 dB at the cutoff frequency; thus, the cutoff is often called the **3-dB frequency** or the *half-power frequency*.

**Table 1 **Correction factors for asymptotic approximation of first-order filter

ω/ω_{0} |
Magnitude response error, (dB) |
Phase response error (deg) |

0.1 | 0 | −5.7 |

0.5 | −1 | 4.9 |

1 | −3 | 0 |

2 | −1 | −4.9 |

10 | 0 | +5.7 |

The case of an ** RC high-pass filter** is analyzed in the same manner as was done for the

\[\begin{matrix}\frac{{{V}_{0}}}{{{V}_{0}}}=\frac{j\omega CR}{1+j\omega CR}=\frac{j(\omega /{{\omega }_{0}})}{1+j(\omega /{{\omega }_{0}})} & {} & {} \\=\frac{(\omega /{{\omega }_{0}})\angle (\pi /2)}{\sqrt{1+{{(\omega /{{\omega }_{0}})}^{2}}}\angle \arctan (\omega /{{\omega }_{0}})} & {} & (7) \\=\frac{\omega /{{\omega }_{0}}}{\sqrt{1+{{(\omega /{{\omega }_{0}})}^{2}}}}\angle \left( \frac{\pi }{2}-\arctan \frac{\omega }{{{\omega }_{0}}} \right) & {} & {} \\\end{matrix}\]

**Figure 2** depicts the Bode plots for **equation 7**, where the horizontal axis indicates the normalized frequency *ω/ω _{0}*. Straight-line asymptotic approximations may again be determined easily at low and high frequencies. The results are very similar to those for the first-order low-pass filter.

For *ω < ω _{0}*, the Bode magnitude approximation intercepts the origin (

**Figure 2** Bode plots for RC high-pass filter. (a) Magnitude response; (b) phase response

The straight- line approximations of the Bode phase plot are:

- For
*ω*< 0.1*ω*, ∠ (V_{o}_{o}/V) ≈ π/2._{i} - 2. For 0.1ω
_{o}and 10ω_{o}, ∠ (V_{o}/V≈ − (π/4) log (10_{i)}*ω*/*ω*)._{o} - For
*ω > 10ω*, ∠ (V_{o}/V_{o}) ≈ 0._{i}

These straight-line approximations are illustrated in **Figure 2(b).**

Bode plots of high-order systems may be obtained by combining Bode plots of factors of the higher-order frequency response function. Let, for example,

\[H (j\omega )={{H }_{1}}(j\omega ){{H }_{2}}(j\omega ){{H }_{3}}(j\omega )\begin{matrix}{} & {} & ( \\\end{matrix}8)\]

which can be expressed, in logarithmic form, as

\[{{\left| H (j\omega ) \right|}_{dB}}={{\left| {{H }_{1}}(j\omega ) \right|}_{dB}}+{{\left| {{H }_{2}}(j\omega ) \right|}_{dB}}+{{\left| {{H }_{3}}(j\omega ) \right|}_{dB}}\begin{matrix}{} & {} & ( \\\end{matrix}9)\]

And

\[\angle H (j\omega )=\angle {{H }_{1}}(j\omega )+\angle {{H }_{2}}(j\omega )+\angle {{H }_{3}}(j\omega )\begin{matrix}{} & {} & ( \\\end{matrix}10)\]

Consider as an example the frequency response function

\[H (j\omega )=\frac{j\omega +5}{(j\omega +10)(j\omega +100)}\begin{matrix}{} & {} & (11) \\\end{matrix}\]

The first step in computing the asymptotic approximation consists of factoring each term in the expression so that it appears in the form a_{i} ( jω/ω_{i} +1), where the frequency ω* _{i}* corresponds to the appropriate 3-dB frequency, ω

\[\begin{matrix}H (j\omega )=\frac{5(j\omega /5+1)}{10(j\omega /10+1)100(j\omega /100+1)} & {} & {} \\{} & {} & (12) \\\frac{0.005(j\omega /5+1)}{10(j\omega /10+1)100(j\omega /100+1)}=\frac{K(j\omega /{{\omega }_{1}}+1)}{(j\omega /{{\omega }_{2}}+1)(j\omega /{{\omega }_{3}}+1)} & {} & {} \\\end{matrix}\]

**Equation 12** can now be expressed in logarithmic form:

\[\begin{matrix}H (j\omega ){{\left| _{dB}=\left| 0.005 \right| \right.}_{dB}}+\left| \frac{j\omega }{5}+1 \right|-\left| \frac{j\omega }{10}+1 \right|-\left| \frac{j\omega }{100}+1 \right| & {} & {} \\{} & {} & (13) \\\angle H (j\omega )=\angle 0.005+\angle \left( \frac{j\omega }{5}+1 \right)-\angle \left( \frac{j\omega }{10}+1 \right)-\angle \left( \frac{j\omega }{5}+1 \right) & {} & {} \\\end{matrix}\]

Each of the terms in the logarithmic **magnitude expression** can be plotted individually.

The constant corresponds to the value −46 dB, plotted in **Figure 3(a)** as a line of zero slope.

The numerator term, with a 3-dB frequency *ω _{1}* = 5, is expressed in the form of the first-order Bode plot of

You see that the individual factors are very easy to plot by inspection once the frequency response function has been normalized in the form of **equation 9.**

If we now consider the **phase response portion** of **equation 13,** we recognize that the first term, the phase angle of the constant, is always zero.

The numerator first-order term, on the other hand, can be approximated, that is, by drawing a straight line starting at 0.1ω_{1} =0.5, with slope +π/4rad/decade *(positive because this is a numerator factor)* and ending at 10ω_{1} = 50, where the asymptote +π/2 is reached.

The two denominator terms have similar behavior, except for the fact that the slope is −π/4 and that the straight line with slope −π/4 rad/decade extends between the frequencies 0.1ω_{2 }and 10ω_{2}, and 0.1ω_{3} and 10ω_{3}, respectively.

**Figure 3** depicts the asymptotic approximations of the individual factors in **equation 13**, with the magnitude factors shown in **Figure 3(a)** and the phase factors in **Figure 3(b).** When all the asymptotic approximations are combined, the complete frequency response approximation is obtained.

**Figure 4** depicts the results of the asymptotic Bode approximation when compared with the actual frequency response functions.

**Figure 3** Bode plot approximation for a second-order frequency response function. (a) Straight-line approximation of magnitude response; (b) straight-line approximation of phase angle response

**Figure 4** Comparison of Bode plot approximation with the actual frequency response function. (a) Magnitude response of second-order frequency response function; (b) phase angle response of second-order frequency response function.

You can see that once a frequency response function is factored into the appropriate form, it is relatively easy to sketch a good approximation of the Bode plot, even for higher-order frequency response functions.

This section illustrates the Bode plot asymptotic approximation construction procedure. The method assumes that there are no complex conjugate factors in the response and that both the numerator and denominator can be factored into first-order terms with real roots.

1. Express the frequency response function in factored form, resulting in an expression similar to the following:

\[H\left( j\omega \right)=\frac{K\left( {j\omega }/{{{\omega }_{1}}+1}\; \right)\cdots \left( {j\omega }/{{{\omega }_{m}}+1}\; \right)}{\left( {j\omega }/{{{\omega }_{m+1}}+1}\; \right)\cdots \left( {j\omega }/{{{\omega }_{n}}+1}\; \right)}\]

2. Select the appropriate frequency range for the semi logarithmic plot, extending at least a decade below the lowest 3-dB frequency and a decade above the highest 3-dB frequency.

3. Sketch the magnitude and phase response asymptotic approximations for each of the first-order factors, using the techniques illustrated in **Figures 1 to 4**.

4. Add, graphically, the individual terms to obtain a composite response.

5. If desired, apply the correction factors of **Table 1**.

The post Low Pass and High Pass Filter Bode Plot appeared first on Electrical A2Z.

]]>The post Band Pass Filter Frequency Response appeared first on Electrical A2Z.

]]>\[H (j\omega )=\frac{{{V}_{0}}}{{{V}_{i}}}(j\omega )\]

**Figure 1** *RLC* bandpass filter.

Apply voltage division to find:

\[\begin{matrix}{{V}_{0}}(j\omega )={{V}_{i}}(j\omega )\frac{R}{R+1/j\omega C+j\omega L} & {} & {} \\{} & {} & (1) \\={{V}_{i}}(j\omega )\frac{j\omega CR}{1+j\omega CR+{{(j\omega )}^{2}}LC} & {} & {} \\\end{matrix}\]

Thus, the frequency response function is:

\[\frac{{{V}_{0}}}{{{V}_{i}}}(j\omega )=\frac{j\omega CR}{1+j\omega CR+{{(j\omega )}^{2}}LC}\begin{matrix}{} & (2) \\\end{matrix}\]

**Equation 2** be factored into the form

\[\frac{{{V}_{0}}}{{{V}_{i}}}(j\omega )=\frac{jA\omega }{(j\omega /{{\omega }_{1}}+1)+(j\omega /{{\omega }_{2}}+1)}\begin{matrix}{} & (3) \\\end{matrix}\]

Where *ω _{1}* and

An immediate observation we can make is that if the signal frequency *ω* is zero, the response of the filter is equal to zero since at *ω* = 0 the impedance of the capacitor 1/*jωC* **becomes infinite**. Thus, the capacitor acts as an **open-circuit**, and the output voltage equals zero.

Further, we note that the filter output in response to an input signal at sinusoidal frequency approaching infinity is again equal to zero. This result can be verified by considering that as ω approaches infinity, the impedance of the **inductor becomes infinite**, that is, an **open-circuit**. Thus, the filter cannot pass signals at very high frequencies.

In an intermediate band of frequencies, the bandpass filter circuit will provide a variable attenuation of the input signal, dependent on the frequency of the excitation. This may be verified by taking a closer look at **equation 1:**

\[\begin{matrix}H (j\omega )=\frac{{{V}_{0}}}{{{V}_{i}}}(j\omega )=\frac{jA\omega }{(j\omega /{{\omega }_{1}}+1)+(j\omega /{{\omega }_{2}}+1)} & {} & {} \\=\frac{A\omega {{e}^{j\pi /2}}}{\sqrt{1+{{(\omega /{{\omega }_{1}})}^{2}}}\sqrt{1+{{(\omega /{{\omega }_{2}})}^{2}}}{{e}^{j\arctan (\omega /{{\omega }_{1}})}}{{e}^{j\arctan (\omega /{{\omega }_{2}})}}} & {} & \left( 4 \right) \\=\frac{A\omega }{\sqrt{\left[ 1+{{(\omega /{{\omega }_{1}})}^{2}} \right]\left[ 1+{{(\omega /{{\omega }_{2}})}^{2}} \right]}}{{e}^{j\left[ \pi /2-\arctan (\omega /{{\omega }_{1}})-\arctan (\omega /{{\omega }_{2}}) \right]}} & {} & {} \\\end{matrix}\]

**Equation 2** is of the form **H** (*jω*) =|H|e^{j∠H}, with

\[\begin{matrix} \begin{align} & \left| H (j\omega ) \right|=\frac{A\omega }{\sqrt{\left[ 1+{{(\omega /{{\omega }_{1}})}^{2}} \right]\left[ 1+{{(\omega /{{\omega }_{2}})}^{2}} \right]}} \\ & \angle H (j\omega )=\frac{\pi }{2}-\arctan \frac{\omega }{{{\omega }_{1}}}-\arctan \frac{\omega }{{{\omega }_{2}}} \\\end{align} & {} & (5) \\\end{matrix}\]

The magnitude and phase plots for the frequency response of the bandpass filter of **Figure 1** are shown in **Figure 2**. These plots have been normalized to have the filter passband centered at the frequency *ω* = 1 rad/s.

The frequency response plots of **Figure 2** suggest that, in some sense, the bandpass filter acts as a combination of a high-pass and a low-pass filter. As illustrated in the previous cases, it should be evident that one can adjust the filter response as desired simply by selecting appropriate values for *L, C*, and *R*.

**Figure 2** Frequency response of RLC bandpass filter

The response of second-order filters can be explained more generally by rewriting the frequency response function of the second-order bandpass filter of **Figure 1** in the following forms:

\[\begin{matrix}\frac{{{V}_{0}}}{{{V}_{_{i}}}}(j\omega )=\frac{j\omega CR}{LC{{(j\omega )}^{2}}+j\omega CR+1} & {} & {} \\=\frac{(2\zeta /{{\omega }_{n}})j\omega }{{{(j\omega /{{\omega }_{n}})}^{2}}+(2\zeta /{{\omega }_{n}})j\omega +1} & {} & (6) \\=\frac{(1/Q{{\omega }_{n}})j\omega }{{{(j\omega /{{\omega }_{n}})}^{2}}+(1/Q{{\omega }_{n}})j\omega +1} & {} & {} \\\end{matrix}\]

with the following definitions:

\[\begin{matrix}{{\omega }_{n}}=\sqrt{\frac{1}{LC}}=natural\begin{matrix}or\begin{matrix}resonant\begin{matrix}frequency \\\end{matrix} \\\end{matrix} \\\end{matrix} & {} & {} \\Q=\frac{1}{2\zeta }=\frac{1}{{{\omega }_{n}}CR}={{\omega }_{n}}\frac{L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}=quality\text{ }factor & {} & (7) \\\zeta =\frac{1}{2Q}=\frac{R}{2}\sqrt{\frac{C}{L}}=damping\text{ }ratio & {} & {} \\\end{matrix}\]

**Figure 3** depicts the normalized frequency response (magnitude and phase) of the second-order band pass filter for *ω _{n}* = 1 and various values of Q (and ζ). The peak displayed in the frequency response around the frequency

Note that as the **quality factor** *Q* increases, the sharpness of the resonance increases and the filter becomes increasingly ** selective** (i.e., it has the ability to filter out most frequency components of the input signals except for a narrow band around the resonant frequency).

One measure of the selectivity of a bandpass filter is its **bandwidth**. The concept of bandwidth can be easily visualized in the plot of **Figure 3(a**) by drawing a horizontal line across the plot (we have chosen to draw it at the amplitude ratio value of 0.707 for reasons that will be explained shortly). The frequency range between (magnitude) frequency response points intersecting this horizontal line is defined as the **half-power bandwidth** of the filter.

The name ** half-power** stems from the fact that when the amplitude response is equal to 0.707 (or ${}^{1}/{}_{\sqrt{2}}$ ), the voltage (or current) at the output of the filter has decreased by the same factor, relative to the maximum value (at the resonant frequency). Since power in an electric signal is proportional to the square of the voltage or current, a drop by a factor ${}^{1}/{}_{\sqrt{2}}$ in the output voltage or current corresponds to the power being reduced by a factor of ½.

Thus, we term the frequencies at which the intersection of the 0.707 line with the frequency response occurs the **half-power frequencies**. Another useful definition of bandwidth *B* is as follows. Note that a high-*Q* filter has a narrow bandwidth and a low-* Q* filter has a wide band with.

\[B=\frac{{{\omega }_{n}}}{Q}\begin{matrix}{} & Ban{{d}_{{}}}width & \begin{matrix}{} & (8) \\\end{matrix} \\\end{matrix}\]

**Figure 3(a)** Normalized magnitude response of second-order bandpass filter; **(b)** normalized phase response of second-order bandpass filter

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]]>**Figure 1** depicts a simple **RC filter** and denotes its input and output voltages, respectively, by *V _{i}* and

**Figure 1** RC Low-pass filter

The frequency response for the filter may be obtained by considering the function

\[H (j\omega )=\frac{{{V}_{0}}}{{{V}_{i}}}\left( j\omega \right)\begin{matrix}{} & (1) \\\end{matrix}\]

and noting that the output voltage may be expressed as a function of the input voltage by means of a voltage divider, as follows

\[{{V}_{0}}(j\omega )={{V}_{i}}(j\omega )\frac{1/j\omega C}{R+1/j\omega C}={{V}_{i}}(j\omega )\frac{1}{1+j\omega RC}\begin{matrix}{} & (2) \\\end{matrix}\]

Thus, the frequency response of the *RC* filter is

\[\frac{{{V}_{0}}}{{{V}_{i}}}(j\omega )=\frac{1}{1+j\omega CR}\begin{matrix}{} & (3) \\\end{matrix}\]

An immediate observation upon studying this frequency response is that if the signal frequency ω is zero, the value of the frequency response function is 1. That is, the filter is passing all the input. Why? To answer this question, we note that at ω = 0, the impedance of the capacitor, 1*/jωC*, becomes infinite. Thus, the capacitor acts as an open-circuit, and the output voltage equals the input:

\[{{V}_{0}}(j\omega =0)={{V}_{i}}(j\omega =0)\begin{matrix}{} & (4) \\\end{matrix}\]

Since a signal at sinusoidal frequency equal to zero is a DC signal, this filter circuit does not in any way affect DC voltages and currents. As the signal frequency increases, the magnitude of the frequency response decreases since the denominator increases with *ω*. More precisely, **equations 5 to 8** describe the magnitude and phase of the frequency response of the RC filter:

\[\begin{matrix}H (j\omega )=\frac{{{V}_{0}}}{{{V}_{i}}}(j\omega )=\frac{1}{1+j\omega CR} & {} & {} \\=\frac{1}{\sqrt{1+{{(\omega CR)}^{2}}}}=\frac{{{e}^{j0}}}{{{e}^{-j\arctan (\omega CR/1)}}} & {} & \left( 5 \right) \\=\frac{1}{\sqrt{1+{{(\omega CR)}^{2}}}}{{e}^{-j\arctan (\omega CR)}} & {} & {} \\\end{matrix}\]

OR

\[H (j\omega )=\left| H (j\omega ) \right|{{e}^{j\angle H(j\omega )}}\begin{matrix}{} & (6) \\\end{matrix}\]

WITH

\[\left| H(j\omega ) \right|=\frac{1}{\sqrt{1+{{(\omega CR)}^{2}}}}=\frac{1}{\sqrt{1+{{(\omega /{{\omega }_{0}})}^{2}}}}\begin{matrix}{} & (7) \\\end{matrix}\]

AND

\[\angle H (j\omega )=-\arctan (\omega CR)=-\arctan \frac{\omega }{{{\omega }_{0}}}\begin{matrix}{} & (8) \\\end{matrix}\]

WITH

\[{{\omega }_{0}}=\frac{1}{RC}\begin{matrix}{} & (9) \\\end{matrix}\]

The simplest way to envision the effect of the filter is to think of the phasor voltage scaled by a factor of |**H**| and shifted by a phase angle ∠**H** by the filter at each frequency, so that the resultant output is given by the phasor with

\[\begin{matrix}{{V}_{0}}=\left| H \right|.{{V}_{i}} & {} & {} \\{} & {} & (10) \\{{\phi }_{0}}=\angle H +{{\phi }_{i}} & {} & {} \\\end{matrix}\]

and where |**H**| and ∠**H** are functions of frequency. The frequency *ω _{0}* is called the

It is customary to represent **H **(*jω*) in two separate plots, representing |**H**| and ∠**H** as functions of *ω*. These are shown in **Figure 2** in normalized form, that is, with |**H**| and ∠**H** plotted versus *ω/ω _{o}*, corresponding to a cutoff frequency

Note that, in the plot, the frequency axis has been scaled logarithmically. This is a common practice in electrical engineering because it enables viewing a very broad range of frequencies on the same plot without excessively compressing the low- frequency end of the plot.

The frequency response plots of **Figure 2** are commonly employed to describe the frequency response of a circuit since they can provide a clear idea at a glance of the effect of a filter on an excitation signal. The cutoff frequency *ω* = 1/*RC* has a special significance in that it represents approximately the point where the filter begins to filter out the higher-frequency signals. The value of |**H** (jω)| at the cutoff frequency is ${}^{1}/{}_{\sqrt{2}}=0.707$ . Note how the cutoff frequency depends exclusively on the values of *R* and *C*. Therefore, one can adjust the filter response as desired simply by selecting appropriate values for *C* and *R*, and therefore one can choose the desired filtering characteristics.

**Figure 2** Frequency response of an RC low-pass filter

Just as a low-pass filter preserves low-frequency signals and attenuates those at higher frequencies, a **high-pass filter** attenuates low-frequency signals and preserves those at frequencies above a cutoff frequency. Consider the high-pass filter circuit shown in **Figure 3**.

**Figure 3** RC High-pass filter

The frequency response is defined as:

\[H (j\omega )=\frac{{{V}_{0}}}{{{V}_{i}}}(j\omega )\]

Voltage division yields:

\[{{V}_{0}}(j\omega )={{V}_{i}}(j\omega )\frac{R}{R+1/j\omega C}={{V}_{i}}(j\omega )\frac{j\omega CR}{1+j\omega CR}\begin{matrix}{} & (11) \\\end{matrix}\]

Thus, the frequency response of the filter is:

\[\frac{{{V}_{0}}}{{{V}_{i}}}(j\omega )=\frac{j\omega CR}{1+j\omega CR}\begin{matrix}{} & (12) \\\end{matrix}\]

Which can be expressed in magnitude-and-phase form by

\[\begin{align}& H (j\omega )=\frac{{{V}_{0}}}{{{V}_{i}}}(j\omega )=\frac{j\omega CR}{1+j\omega CR}=\frac{\omega CR{{e}^{j\pi /2}}}{\sqrt{1+{{(\omega CR)}^{2}}}{{e}^{j\arctan (\omega CR/1)}}} \\& =\frac{\omega CR}{\sqrt{1+{{(\omega CR)}^{2}}}}.{{e}^{j\left[ \pi /2-\arctan (\omega CR) \right]}} \\\end{align}\]

Or

\[\begin{matrix}H (j\omega )=\left| H \right|{{e}^{j\angle H }} & {} & \left( 13 \right) \\\end{matrix}\]

with

\[\begin{matrix}\left| H (j\omega ) \right|=\frac{\omega CR}{\sqrt{1+{{(\omega CR)}^{2}}}} & {} & {} \\{} & {} & (14) \\\angle H (j\omega )={{90}^{o}}-\arctan (\omega CR) & {} & {} \\\end{matrix}\]

You can verify by inspection that the amplitude response of the high-pass filter will be zero at *ω* = 0 and will asymptotically approach 1 as *ω* approaches infinity while the phase shift is *π*/2 at *ω* = 0 and tends to zero for increasing ω.

Amplitude and phase response curves for the high-pass filter are shown in **Figure 4.** These plots have been normalized to have the filter cutoff frequency *ω*_{0} = 1 rad/s. Note that, once again, it is possible to define a cutoff frequency at *ω _{0}* = 1/

**Figure 4** Frequency response of an RC high-pass filter

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]]>\[\begin{matrix}x(t)=x(t+T)=x(t+nT) & n=1,2,3,…..;T=period & \begin{matrix}{} & (1) \\\end{matrix} \\\end{matrix}\]

An example of a periodic signal is shown in **Figure 1.**

**Figure 1** Thevenin equivalent source network

The signal *x (t)* can be expressed as an infinite summation of sinusoidal components, known as a **Fourier **series, using either of the following two representations.

**Fourier Series:**

- Sine-cosine (quadrature) representation

\[x(t)={{a}_{0}}+\sum\limits_{n=1}^{\infty }{{{a}_{n}}\cos \left( n\frac{2\pi }{T}t \right)+\sum\limits_{n=1}^{\infty }{{{b}_{n}}\sin \left( n\frac{2\pi }{T}t \right)}}\begin{matrix}{} & (2) \\\end{matrix}\]

- Magnitude and phase form

\[x(t)={{c}_{0}}+\sum\limits_{n=1}^{\infty }{{{c}_{n}}}\sin \left( n\frac{2\pi }{T}t+{{\theta }_{n}} \right)\begin{matrix}{} & (3) \\\end{matrix}\]

\[x(t)={{c}_{0}}+\sum\limits_{n=1}^{\infty }{{{c}_{n}}\cos \left( n\frac{2\pi }{T}t-{{\psi }_{n}} \right)}\begin{matrix}{} & (4) \\\end{matrix}\]

In each of these expressions, the period T is related to the **fundamental frequency** of the signal ω_{o} by

\[{{\omega }_{0}}=2\pi {{f}_{0}}=\frac{2\pi }{T}\begin{matrix}{} & rad/s\begin{matrix}{} & (5) \\\end{matrix} \\\end{matrix}\]

The frequencies 2ω_{o}, 3ω_{o}, 4ω_{o}, etc., are called harmonics.

It is straightforward to show that **equations 2 and 3** are equivalent when the parameters *a _{n}, b_{n}, c_{n}* and

\[\begin{matrix}\sqrt{{{a}_{n}}^{2}+{{b}_{n}}^{2}}={{c}_{n}} & and & \frac{{{b}_{n}}}{{{a}_{n}}}=\cot \left( {{\theta }_{n}} \right) & (6) \\\end{matrix}\]

Similarly, one can show that **equations 2 and 4** are equivalent when the parameters *a _{n}, b_{n}, c_{n}*, and

\[\begin{matrix}\sqrt{{{a}_{n}}^{2}+{{b}_{n}}^{2}}={{c}_{n}} & and & \frac{{{b}_{n}}}{{{a}_{n}}}=\tan \left( {{\psi }_{n}} \right) & (7) \\\end{matrix}\]

**Figure 2** is a graphical representation of the equivalence of the {*a _{n}, b_{n}*} and {

**Figure 2** Relationship between {*a _{n }, b_{n}*} and {

Each of the two forms of the Fourier series, **equations 2 and 2 (or 4)**, has its distinct advantages. The sine-cosine representation uses odd and even functions of the independent variable. Odd functions are anti-symmetric about the origin and satisfy the condition:

\[f(-t)=-f(t)\begin{matrix}{} & odd\begin{matrix}function\begin{matrix}{} & (8) \\\end{matrix} \\\end{matrix} \\\end{matrix}\]

Sine functions are odd. Even functions are symmetric about the origin and satisfy the condition:

\[f(-t)=f(t)\begin{matrix}{} & even\begin{matrix}function\begin{matrix}{} & (9) \\\end{matrix} \\\end{matrix} \\\end{matrix}\]

Cosine functions are even, as is any constant, such as *a _{0}*.

**Figure 3** Definition of even and odd functions

The advantage of the representation is that if *x(t)* is known to be odd (even), it can be represented as the sum of only odd (even) functions [i.e., using only the sine (cosine) terms], thus resulting in easier evaluation of the Fourier series coefficients.

The magnitude and phase forms separate out the magnitude information c_{n} from the phase information 𝜃_{n} or *ψ*_{n}. In this form, Fourier series may be combined readily with magnitude and phase representations of linear systems to periodic inputs. The magnitude and phase components are often represented as a discrete **frequency spectrum**, as shown in **Figure 4.**

**Figure 4** Discrete frequency spectrum

The computation of the {*a _{n}, b_{n}*} and {

\[\begin{matrix}{{a}_{0}}=\frac{1}{T}\int\limits_{0}^{T}{x(t)dt}=\frac{1}{T}\int\limits_{-T/2}^{T/2}{x(t)dt} & \text{average value of x(t)} & (9) \\\end{matrix}\]\[\begin{matrix}{{a}_{n}}=\frac{2}{T}\int\limits_{0}^{T}{x(t)\cos \left( n\frac{2\pi }{T}t \right)}dt=\frac{2}{T}\int\limits_{-T/2}^{T/2}{x(t)\cos \left( n\frac{2\pi }{T}t \right)dt} & {} & (10) \\\end{matrix}\]

\[\begin{matrix}{{b}_{n}}=\frac{2}{T}\int\limits_{0}^{T}{x(t)\sin \left( n\frac{2\pi }{T}t \right)}dt=\frac{2}{T}\int\limits_{-T/2}^{T/2}{x(t)\sin \left( n\frac{2\pi }{T}t \right)dt} & {} & (11) \\\end{matrix}\]

The integral limits are written in two different forms to illustrate that it does not matter where the integration starts, provided that it is carried out over one entire period. The c_{n} and 𝜃* _{n}* (or

To illustrate the significance of the Fourier series decomposition, consider the square wave of **Figure 5(a),** which is an even function. Only even function (cosine) terms are non-zero. The first six non-zero Fourier series terms are shown in **Figure 5(b).**

Note that the first term corresponds to *a _{o}*, which is the average value

**Figure 5(c)** compares the original square wave with the Fourier series approximation. It should be evident that 10 terms are not sufficient to reproduce the sharp edges of the square wave, but it should also be clear that as we add more Fourier terms, the resulting approximation will be closer and closer to the square wave signal.

**Figure 5** Square wave and its representation by a Fourier series. (a) Square wave (even function); (b) first six Fourier’s series terms of square ware; (c) sum of first six Fourier series terms superimposed upon a square wave

The complete Fourier series for the square wave shown in **Figure 5(a)** is:

\[f(t)=\frac{A}{2}+\frac{2A}{\pi }\sum\limits_{n=1}^{\infty }{{{(-1)}^{(n-1)}}\frac{\cos \left[ \left( 2n-1 \right)\left( 2\pi \right)t/T \right]}{2n-1}}\]

It is interesting to note that if the square wave shown in **Figure 5(a**) were shifted to the right by *T*/4 the resulting waveform would be an odd function. For such a square wave, its value over the interval *0 < t < T /2 is A* while its value over the interval *T /2 < t < T* is zero. The complete Fourier series now contains only odd functions (sines) and is:

\[f(t)=\frac{A}{2}+\frac{2A}{\pi }\sum\limits_{n=1}^{\infty }{\frac{\sin \left[ \left( 2n-1 \right)\left( 2\pi \right)t/T \right]}{2n-1}}\]

The Fourier series for a few other common waveforms are listed below. In each case the peak-to-trough amplitude is A, the period is *T*, and the average value is *A/2*.

\[f(t)=\frac{A}{2}+\frac{A}{\pi }\sum\limits_{n=1}^{\infty }{\frac{\sin \left[ \left( 2n\pi t/T \right) \right]}{n}}\begin{matrix}{} & \begin{matrix}Sawtooth\begin{matrix}wave \\\end{matrix} \\\end{matrix} \\\end{matrix}\]

\[f(t)=\frac{A}{2}-\frac{4A}{{{\pi }^{2}}}\sum\limits_{n=1}^{\infty }{\frac{\cos \left[ \left( 2n-1 \right)\left( 2\pi \right)t/T \right]}{{{(2n-1)}^{2}}}}\begin{matrix}{} & Triangle\text{ }\begin{matrix}wave \\\end{matrix} \\\end{matrix}\]

These Fourier series converge everywhere that the function itself is differentiable. The Fourier series for the square wave does not converge *at t = 0, T /2, T* . . . while the Fourier series for the sawtooth wave does not converge at *t = 0, T, 2T*…

**Response of Linear Systems to Periodic Inputs**

The frequency response concept is particularly useful when one deals with a system excited by a periodic input, which can be modeled by a Fourier series of sinusoids of known amplitude and phase, but different frequencies. Assume that the Fourier series has a finite number of terms:

\[x(t)={{c}_{0}}+\sum\limits_{n=1}^{N}{{{c}_{n}}\sin \left( n\frac{2\pi }{T}t+{{\theta }_{n}} \right)}\begin{matrix}{} & (12) \\\end{matrix}\]

Each of the *N* sinusoids is characterized by amplitude *c _{n}*, phase

**Figure 6** illustrates the input-output representation of a system, making use of the frequency response concept.

**Figure 6** Response of a linear system to a phasor input

The figure shows that if the input *q _{in} (t)* to a linear system can be represented by the phasor

In the case of a periodic input expressed in terms of a (truncated) Fourier series, we must recognize that each of the input sinusoidal components propagates through the system according to the frequency response. Thus, the discrete **magnitude spectrum of the periodic output signal** in the steady state is equal to the discrete magnitude spectrum of the input signal multiplied by the amplitude ratio of the frequency response of the system at the appropriate frequencies.

The **phase spectrum of the output signal** in the steady state is equal to the phase spectrum of the input signal added to the phase angle frequency response of the system at the appropriate frequencies.

If *x(t)* is the input to a linear system in the form given, and if the linear system has a frequency response function **H**( *jω*), then the output of the system *y(t)* is given by

\[y(t)=\sum\limits_{n=1}^{N}{\left. \left| H \left( j{{\omega }_{n}} \right) \right. \right|}{{\begin{matrix} c \\\end{matrix}}_{n}}\sin \left[ {{\omega }_{n}}t+{{\theta }_{n}}+\angle H \left( j{{\omega }_{n}} \right) \right]\begin{matrix} {} & (13) \\\end{matrix}\]

Where |**H**(jω_{n})| and ∠**H(**jω_{n}) are the magnitude and phase, respectively, at the frequency corresponding to the *n*th harmonic nω_{o} of the input.

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]]>Since second-order circuits have two irreducible storage elements, such circuits have two state variables and their behavior is described by a second-order differential equation.

The simplest, yet arguably the most crucial, second-order circuits are those in which the capacitor and inductor are either in parallel or in series, as shown in **Figures 1 and 2.** The circuits in these figures are drawn to suggest that the capacitor and inductor should be treated as a unified load. The rest of each circuit is either the **Thevenin** or **Norton** equivalent of the source network.

The analysis of these circuits is somewhat less complicated than for other second-order circuits, which appealing for anyone learning to analyze such circuits for the first time.

**General Characteristics**

Before diving into the analysis of particular second-order circuits it is worthwhile to introduce the generalized differential equation for any second-order circuit.

$\begin{matrix}\frac{1}{\omega _{n}^{2}}\frac{{{d}^{2}}x}{d{{t}^{2}}}+\frac{2\zeta }{{{\omega }_{n}}}\frac{dx}{dt}+x={{K}_{s}}f\left( t \right) & {} & \left( 1 \right) \\\end{matrix}$

The constants *ω _{n}* and ζ are the

As will be shown, second-order circuits have three distinct possible responses:

overdamped,critically damped, andunderdamped. The response for any particular second-order circuit is determined entirely by ζ.

In **equation 1**, *f (t )* is a forcing function. *K _{S}* is the

**Parallel LC Circuits**

Consider the circuit shown in **Figure 1.** The two state variables are *i _{L}* and

In general, at the moment of a transient event the energy of the storage elements may be non-zero; that is, the voltage *v _{C} *(0) across the capacitor and the current

$\begin{matrix}{{v}_{C}}\left( {{0}^{+}} \right)={{v}_{C}}\left( {{0}^{-}} \right) & and & {{i}_{L}}\left( {{0}^{+}} \right)={{i}_{L}}\left( {{0}^{-}} \right) \\\end{matrix}$

**Figure 1** Second-order circuit with the inductor and capacitor in parallel acting as a unified load attached to a Norton equivalent network

Apply KCL to either node to find a first-order differential equation in terms of both state variables.

$\begin{matrix}{{I}_{N}}-\frac{{{v}_{C}}}{{{R}_{N}}}-{{i}_{L}}-{{i}_{C}}=0 & {} & KCL \\\end{matrix}$

The KCL equation can be transformed into a second-order differential equation in *i _{L}* by recognizing that:

$\begin{matrix}{{v}_{C}}={{v}_{L}}=L\frac{d{{i}_{L}}}{dt} & and & {{i}_{C}} \\\end{matrix}=C\frac{d{{v}_{C}}}{dt}=LC\frac{{{d}^{2}}{{i}_{L}}}{d{{t}^{2}}}$

Substitute for *v _{C}* and

${{I}_{N}}-\frac{L}{{{R}_{N}}}\frac{d{{i}_{L}}}{dt}-{{i}_{L}}-LC\frac{{{d}^{2}}{{i}_{L}}}{d{{t}^{2}}}=0$

Rearrange the order of terms to yield:

$LC\frac{{{d}^{2}}{{i}_{L}}}{d{{t}^{2}}}+\frac{L}{{{R}_{N}}}\frac{d{{i}_{L}}}{dt}+{{i}_{L}}={{I}_{N}}\begin{matrix}{} & \left( 2 \right) & {} \\\end{matrix}$

Alternatively, one can differentiate both sides of the KCL equation and substitute:

$\begin{matrix}\frac{d{{i}_{L}}}{dt}=\frac{{{v}_{C}}}{L} & and & \frac{d{{i}_{C}}}{dt}=C\frac{{{d}^{2}}{{v}_{C}}}{d{{t}^{2}}} \\\end{matrix}$

The result is:

$\frac{d{{I}_{N}}}{dt}-\frac{1}{{{R}_{N}}}\frac{d{{v}_{C}}}{dt}-\frac{{{v}_{C}}}{L}-C\frac{{{d}^{2}}{{v}_{C}}}{d{{t}^{2}}}=0$

Multiply both sides of the equation by L, and if the source IN is a constant such that its time derivative is zero, the resulting second-order differential equation is:

$\begin{matrix}LC\frac{{{d}^{2}}{{v}_{C}}}{d{{t}^{2}}}+\frac{L}{{{R}_{N}}}\frac{d{{v}_{C}}}{dt}+{{v}_{C}}=0 & {} & \left( 3 \right) \\\end{matrix}$

**Equation 3** contains no forcing function (its right side is zero), which indicates that the long-term steady-state solution for *v _{C}* will be zero. In other words, the transient solution for

To solve **equations 2 and 3** it is first necessary to identify *the dimension – less damping ratio* ζ and the *natural frequency* *ω _{n}*. Notice that the left sides of both equations are identical, as they are for any variable in the circuit. Thus,

$\begin{matrix}\frac{1}{\omega _{n}^{2}}=LC & and & \frac{2\zeta }{{{\omega }_{n}}} \\\end{matrix}=\frac{L}{{{R}_{N}}}$

These two equations can be solved to yield:

$\begin{matrix}{{\omega }_{n}}=\frac{1}{\sqrt{LC}} & and & \zeta =\frac{1}{2{{R}_{N}}}\sqrt{\frac{L}{C}} & \left( 4 \right) \\\end{matrix}$

The type of transient response for *i _{L}* and

${{i}_{L}}\left( t \right)={{\left( {{i}_{L}} \right)}_{TR}}+{{\left( {{i}_{L}} \right)}_{SS}}={{\left( {{i}_{L}} \right)}_{TR}}+{{I}_{N}}$

And

${{v}_{C}}\left( t \right)={{\left( {{v}_{C}} \right)}_{TR}}+{{\left( {{v}_{C}} \right)}_{SS}}={{\left( {{v}_{C}} \right)}_{TR}}+L\frac{d{{I}_{N}}}{dt}$

Note that when *I _{N}* is a constant, (

**Series LC Circuits**

The development of the general solution for series *LC* circuits follows the same basic steps used above for parallel *LC *circuits. Consider the circuit in **Figure 2** and note the duality between what follows and what was done above for the parallel *LC* circuit. In fact, the equations that follow can be found simply by starting with the equations developed above and swapping *L* with *C*, i_{L} with v_{C} , R_{N} with 1/R_{T} , and I_{N} with V_{T.}

**Figure 2** Second-order circuit with the inductor and capacitor in series acting as a unified load attached to a Thevenin equivalent network

Again, the two state variables are i_{L} and v_{C} , where i_{L} is the primary state variable because it is shared by all four circuit elements. Again, at the moment of a transient event the energy of the storage elements may be non-zero; that is, the voltage *v _{C}* (0) across the capacitor and the current i

$\begin{matrix}{{v}_{C}}\left( {{0}^{+}} \right)={{v}_{C}}\left( {{0}^{-}} \right) & and & {{i}_{L}}\left( {{0}^{+}} \right) \\\end{matrix}={{i}_{L}}\left( {{0}^{-}} \right)$

Apply KVL around the series loop to find a first-order differential equation in terms of both state variables.

$\begin{matrix}{{V}_{T}}-{{i}_{L}}{{R}_{T}}-{{v}_{C}}-{{v}_{L}}=0 & {} & KVL \\\end{matrix}$

The KVL equation can be transformed into a second-order differential equation in *v _{C} *by recognizing that:

\[\begin{matrix}{{i}_{L}}={{i}_{C}}=C\frac{d{{v}_{C}}}{dt} & and & {{v}_{L}}=L\frac{d{{i}_{L}}}{dt}=LC\frac{{{d}^{2}}{{v}_{C}}}{d{{t}^{2}}} \\\end{matrix}\]

Substitute for *v _{L}* and

\[{{V}_{T}}-{{R}_{T}}C\frac{d{{v}_{C}}}{dt}-{{v}_{C}}-LC\frac{{{d}^{2}}{{v}_{C}}}{d{{t}^{2}}}=0\]

Rearrange the order of terms to yield:

$\begin{matrix}LC\frac{{{d}^{2}}{{v}_{C}}}{d{{t}^{2}}}+{{R}_{T}}C\frac{d{{v}_{C}}}{dt}+{{v}_{C}}={{V}_{T}} & {} & \left( 5 \right) \\\end{matrix}$

Alternatively, one can differentiate both sides of the KVL equation and substitute:

$\begin{matrix}\frac{d{{v}_{C}}}{dt}=\frac{{{i}_{L}}}{C} & and & \frac{d{{v}_{L}}}{dt}=L\frac{{{d}^{2}}{{i}_{L}}}{d{{t}^{2}}} \\\end{matrix}$

The result is:

$\frac{d{{V}_{T}}}{dt}-{{R}_{_{T}}}\frac{d{{i}_{L}}}{dt}-\frac{{{i}_{L}}}{C}-L\frac{{{d}^{2}}{{i}_{L}}}{d{{t}^{2}}}=0$

Multiply both sides of the equation by *C*, and if the source V_{T} is a constant such that its time derivative is zero, the resulting second-order differential equation is:

$\begin{matrix}LC\frac{{{d}^{2}}{{i}_{L}}}{d{{t}^{2}}}+{{R}_{T}}C\frac{d{{i}_{L}}}{dt}+{{i}_{L}}=0 & {} & \left( 6 \right) \\\end{matrix}$

**Equation 6** contains no forcing function (its right side is zero), which indicates that the long-term steady-state solution for i_{L} will be zero. In other words, the transient solution for i_{L} is also its complete solution.

To solve **equations 5 and 6** it is first necessary to identify the *dimensionless damping* ratio ζ and the *natural frequency ω _{n}*. Notice that the left sides of both equations are identical, as they are for any variable in the circuit. Thus,

$\frac{1}{\omega _{n}^{2}}=LC\begin{matrix}{} & and & \frac{2\zeta }{{{\omega }_{n}}} \\\end{matrix}={{R}_{T}}C$

These two equations can be solved to yield:

$\begin{matrix}{{\omega }_{n}}=\frac{1}{\sqrt{LC}} & and & \zeta =\frac{{{R}_{T}}}{2} \\\end{matrix}\sqrt{\frac{C}{L}}\begin{matrix}{} & {} & \left( 7 \right) \\\end{matrix}$

The type of transient response for* i _{L}* and

${{v}_{C}}\left( t \right)={{\left( {{v}_{C}} \right)}_{TR}}+{{\left( {{v}_{C}} \right)}_{SS}}={{\left( {{v}_{C}} \right)}_{TR}}+{{V}_{T}}$

And

${{i}_{L}}\left( t \right)={{\left( {{i}_{L}} \right)}_{TR}}+{{\left( {{i}_{L}} \right)}_{SS}}={{\left( {{i}_{L}} \right)}_{TR}}+C\frac{d{{V}_{T}}}{dt}$

Note that when *V _{T}* is a constant,

The transient response *x _{TR}(t)* is found by setting the right side of the governing differential equation equal to zero. That is:

$\begin{matrix}\frac{1}{\omega _{n}^{2}}\frac{{{d}^{2}}{{x}_{TR}}}{d{{t}^{2}}}+\frac{2\zeta }{{{\omega }_{n}}}\frac{d{{x}_{TR}}}{dt}+{{x}_{TR}}=0 & {} & \left( 8 \right) \\\end{matrix}$

Just as in first-order systems, the solution of this equation has an exponential form:

${{x}_{TR}}\left( t \right)=\alpha {{e}^{st}}\begin{matrix}{} & \left( 9 \right) & {} \\\end{matrix}$ **Assumed transient response**

Substitution into the differential equation yields the *characteristic equation:*

$\frac{{{s}^{2}}}{\omega _{n}^{2}}+\frac{2\zeta }{{{\omega }_{n}}}+1=0\begin{matrix} {} & {} & \left( 10 \right) \\\end{matrix}$

which, in turn, yields two **characteristic roots** s_{1} and s_{2}. Specific values of s1 and s2 are found from the quadratic formula applied to the characteristic equation:

$\begin{matrix}{{s}_{1,2}}=-\zeta {{\omega }_{n}}\pm \frac{1}{2}\sqrt{{{\left( 2\zeta {{\omega }_{n}} \right)}^{2}}-4\omega _{n}^{2}}=-{{\omega }_{n}}\left( \zeta \pm \sqrt{{{\zeta }^{2}}-1} \right) & {} & \left( 11 \right) \\\end{matrix}$

The roots s_{1 }and s_{2} are associated with the three distinct possible responses: over- damped (ζ > 1), critically damped (ζ = 1), and underdamped (ζ < 1). The details of each of these responses are presented below.

**Overdamped Response (ζ > 1)**

**Two distinct, negative and real roots: (s _{1}, s_{2}).**The transient response is over-damped when ζ > 1 and the roots are${{s}_{1,2}}={{\omega }_{n}}(-\zeta \pm \sqrt{{{\zeta }^{2}}-1})$. The general form of the solution is:

${{x}_{TR}}\left( t \right)={{\alpha }_{1}}{{e}^{{{s}_{1}}t}}+{{\alpha }_{2}}{{e}^{{{s}_{2}}t}}={{e}^{-\zeta {{\omega }_{n}}t}}\left[ {{\alpha }_{1}}{{e}^{\left( {{\omega }_{n}}\sqrt{{{\zeta }^{2}}-1} \right)t}}+{{\alpha }_{2}}{{e}^{\left( -{{\omega }_{n}}\sqrt{{{\zeta }^{2}}-1} \right)t}} \right]$

$={{\alpha }_{1}}{{e}^{-t/{{\tau }_{1}}}}+{{\alpha }_{2}}{{e}^{-t/{{\tau }_{2}}}}$

$\begin{matrix}{{\tau }_{1}}=\frac{1}{{{\omega }_{n}}\left( \zeta -\sqrt{{{\zeta }^{2}}-1} \right)} & {} & {{\tau }_{2}}=\frac{1}{{{\omega }_{n}}\left( \zeta +\sqrt{{{\zeta }^{2}}-1} \right)} & (12) \\\end{matrix}$

Thus, an overdamped response is the sum of two first-order responses, as shown in **Figure 3.**

**Figure 3** Transient response of underdamped second-order system

**Critically Damped Response (ζ = 1)**

**Two identical, negative, and real roots:** (s_{1}, s_{2}). The transient response is critically damped when ζ = 1. The general form of the solution is:

${{x}_{TR}}\left( t \right)={{\alpha }_{1}}{{e}^{{{s}_{1}}t}}+{{\alpha }_{2}}t{{e}^{{{s}_{2}}t}}={{e}^{-{{\omega }_{n}}t}}\left( {{\alpha }_{1}}+{{\alpha }_{2}}t \right)={{e}^{-t/\tau }}\left( {{\alpha }_{1}}+{{\alpha }_{2}}t \right)$

$\begin{matrix}\tau =\frac{1}{{{\omega }_{n}}} & {} & \left( 13 \right) \\\end{matrix}$

Note that a critically damped response is the sum of a first-order exponential term plus a similar term multiplied by *t*. These two components and the complete response are shown in **Figure 4.**

**Figure 4** Transient response of a critically damped second-order system

**Underdamped Response (ζ < 1)**

**Two complex conjugate roots: **(s_{1}, s_{2}). The transient response is underdamped when ζ < 1. The argument of the square root is negative in **equation 11**, such that ${{s}_{1,2}}={{\omega }_{n}}\left( -\zeta \pm j\sqrt{1-{{\zeta }^{2}}} \right)$ . The following complex exponentials appear in the general form of the response:

$\begin{matrix}{{e}^{^{{{\omega }_{n}}}\left( -\zeta +j\sqrt{1-{{\zeta }^{2}}} \right)t}} & {{e}^{^{{{\omega }_{n}}}\left( -\zeta -j\sqrt{1-{{\zeta }^{2}}} \right)t}} & \left( 14 \right) \\\end{matrix}$

Euler’s formula can be used to express the complex exponentials in terms of sinusoids. The result is:

$\begin{matrix}{{x}_{TR}}\left( t \right)={{e}^{-\zeta {{\omega }_{n}}t}}\left[ {{\alpha }_{1}}\sin \left( {{\omega }_{d}}t \right)+{{\alpha }_{2}}\cos \left( {{\omega }_{d}}t \right) \right] & {} & \left( 15 \right) \\\end{matrix}$

Where ${{\omega }_{d}}={{\omega }_{n}}\sqrt{1-{{\zeta }^{2}}}$ is the **damped natural frequency**. Note that *ω _{d}* is the frequency of oscillation and is related to the period T of oscillation by

**Figure 5** Transient response of an underdamped second-order system for α_{1} = α_{2} = 1; ζ = 0.2; ω_{n} = 1

For switched DC sources, the forcing function *F* in equation 5.40 is a constant. The result is a constant long-term *(t → ∞)* steady-state response x_{SS.}

$\begin{matrix}\frac{1}{\omega _{n}^{2}}\frac{{{d}^{2}}{{x}_{SS}}}{d{{t}^{2}}}+\frac{2\zeta }{{{\omega }_{n}}}\frac{d{{x}_{SS}}}{dt}+{{x}_{SS}}={{K}_{s}}F & {} & \left( 16 \right) \\\end{matrix}$

Since x* _{SS}* must also be a constant the solution for x

${{x}_{SS}}=x\left( \infty \right)={{K}_{s}}F\begin{matrix}{} & t\to \infty & \left( 17 \right) \\\end{matrix}$

As with **first-order systems**, the complete response is the sum of the transient and long-term steady-state responses. The complete mathematical solutions for the over- damped, critically damped, and underdamped cases. In each of these cases, the initial conditions on the storage elements must be used to solve for the unknown constants *α _{1}* and

One particularly useful complete solution is the *unit-step response* brought about by letting *K _{S}f (t)* be a unit step, which equals 0 for

**Figure 6** Second order unit step response K_{S} = 1, ω_{d} = 1, and ζ = 0.1

Also, as seen in the underdamped transient response, the magnitude of the oscillation’s decays exponentially over time. The time constant for the surrounding envelope.

Note that the rate of decay of the oscillations is governed by ζ. **Figure 7** shows that as ζ increases the *overshoot* of the long-term DC steady-state response decreases until, when ζ = 1 (critically damped), the response no longer oscillates and the overshoot is zero. The response for ζ < 1 (overdamped) has no oscillations and zero overshoot.

**Figure 7** Second-order unit step response K_{S} = 1, ω_{d} = 1, and ζ ranging from 0.2 to 4.0

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]]>The post First Order System Transient Response appeared first on Electrical A2Z.

]]>Ideal first-order electrical systems possess either capacitance or inductance (but not both) along with resistance and (perhaps) energy sources. Ideal first-order mechanical systems possess mass and damping (e.g., sliding or viscous friction) but no elasticity or compliance.

An ideal first-order fluid system possesses fluid capacitance and viscous dissipation, such as a hydraulic system with a liquid-filled tank and a variable orifice. Many conductive and convective thermal systems also exhibit first-order behavior.

In general, when solving transient circuit problems, it is necessary to determine **three elements**: **(1)** the steady-state response prior to a transient event, **(2)** the transient response immediately following the transient event, and **(3)** the long-term steady-state response remaining after the transient response has decayed away.

The steps involved in computing the complete response of a first-order circuit with constant sources are outlined below. The methodology is straightforward and can be mastered with only modest practice.

**Circuit Simplification for t > 0**

The first step to solve for the response after the transient event (*t* > 0) is to partition the circuit into a source network and load, with the energy storage element as the load, as shown in **Figure 1.** If the source network is linear, it can be replaced by its The ́venin or Norton equivalent network.

**Figure 1** Generalized first-order circuit seen as a source network attached to an energy storage element as the load

Consider the case when the load is a capacitor and the source network is replaced by its **Thevenin equivalent network**, as shown in **Figure 2**.

**Figure 2** Generalized first-order circuit with a capacitor load and a Thevenin source

KVL can be applied around the loop to yield:

${{V}_{T}}-i{{R}_{T}}-{{v}_{C}}=0$

Of course, *i* = *i _{C}* and for a capacitor

$\begin{matrix}{{R}_{T}}C\frac{d{{v}_{C}}}{dt}+{{v}_{C}}={{V}_{T}} & Capacitor\text{ }load\text{ }with\text{ }Thevenin\text{ }source & \left( 1 \right) \\\end{matrix}$

For a DC source network, the long-term steady-state solution is simply v_{C} = V_{T}.

Likewise, consider the case when the load is an inductor and the source network is replaced by its Norton equivalent network, as shown in **Figure 3.**

**Figure 3** Generalized first-order circuit with an inductor load and a Norton source

KCL can be applied at either node to yield.

\[{{I}_{N}}-\frac{v}{{{R}_{N}}}-i=0\]

Of course, *v* = v_{L} and for an inductor v* _{L}* = Ldi

$\begin{matrix}\frac{L}{{{R}_{N}}}\frac{d{{i}_{L}}}{dt}+{{i}_{L}}={{I}_{N}} & Inductor\text{ }load\text{ }with\text{ }Norton\text{ }source & \left( 2 \right) \\\end{matrix}$

For a DC source network, the long-term steady-state solution is simply *i _{L} = I_{N}* . It is also possible to use a Norton or The ́venin source when the load is a capacitor or inductor, respectively. The results can be shown to be identical to those found above by substituting the source transformation expressions

It is important to keep in mind that these solutions are for *t* > 0, that is, the transient response. It is possible that the equivalent source network seen by the load after the transient event is different from that seen by the load before the event. Equivalent network methods can be used for both domains but do not assume that the equivalent network seen by the load is unchanged by the event.

**First-Order Differential Equation**

Both **equations 1 and 2** have the same general form:

$\begin{matrix}\tau \frac{dx\left( t \right)}{dt}+x\left( t \right)={{K}_{S}}f\left( t \right)\text{ }t\ge 0\text{ } & First-order\text{ }system\text{ }equation & \left( 3 \right) \\\end{matrix}$

Where the constants *τ* and *K _{S}* are the

$\begin{matrix}\tau \frac{dx\left( t \right)}{dt}+x\left( t \right)={{K}_{S}}F\left( t \right) & t\ge 0 & \left( 4 \right) \\\end{matrix}$

The solution for *x(t)* has two parts: the transient response and the *long-term steady-state* response. These two parts can also be rearranged in terms of **natural** and **forced** responses. The sum of both parts is known as the **complete response**. One initial condition $x\left( {{0}^{+}} \right)$is needed to specify the complete response.

The transient response *x _{TR}* is found by setting F = 0 in

$\begin{matrix}\tau \frac{d{{x}_{T}}_{R}\left( t \right)}{dt}+{{x}_{TR}}\left( t \right)=0 & {} & \left( 5 \right) \\\end{matrix}$

The solution for x is found by assuming a solution of the form:

$\begin{matrix}{{x}_{TR}}\left( t \right)=\alpha {{e}^{st}} & {} & \left( 6 \right) \\\end{matrix}$

Substitution of this assumed solution into **equation 5** results in a characteristic equation:

$\begin{matrix}\tau s+1=0 & Characrteristic\_equation & \left( 7 \right) \\\end{matrix}$

The solution for s is simply:

$\begin{matrix}s=\frac{-1}{\tau } & {} & \left( 8 \right) \\\end{matrix}$

which is known as the root of the characteristic equation. Plugging in for s in **equation 6** yields a decaying exponential.

$\begin{matrix} {{x}_{TR}}\left( t \right)=\alpha {{e}^{-t/\tau }} & \text{Transient Response} & \left( 9 \right) \\\end{matrix}$

The constant α in **equation 9** cannot be evaluated until the complete response has been found. If the system does not have an external forcing function, the transient response is also the complete response, and the constant α is equal to the initial condition x(0^{+}).

The amplitude of *x _{TR} (t)* at

**Figure 4** Normalized first-order exponential decay

Still assuming that the first-order circuit contains only DC sources, such that *f (t )* is a constant *F*, the long-term steady-state response of a first-order system is the solution to:

$\begin{matrix}\tau \frac{d{{x}_{ss}}\left( t \right)}{dt}={{x}_{ss}}={{K}_{s}}F & t\ge 0 & \left( 10 \right) \\\end{matrix}$

For constant *F*, *x _{SS}* =

$\begin{matrix}{{x}_{ss}}\left( t \right)\equiv x\left( \infty \right)={{K}_{s}}F & \text{F=Constant} & \left( 11 \right) \\\end{matrix}$

The complete response is the sum of the transient and long-term steady-state responses:

$x\left( t \right)={{x}_{TR}}\left( t \right)+{{x}_{SS}}\left( t \right)=\alpha {{e}^{-t/\tau }}+x\left( \infty \right)=\alpha {{e}^{-t/\tau }}+KsF\begin{matrix}{} & t\ge 0 & \left( 12 \right) \\\end{matrix}$

Apply the one initial condition *x*(0^{+}) to solve for the unknown constant α:

$\begin{matrix}x\left( {{0}^{+}} \right)=\alpha +x\left( \infty \right) & {} & {} \\{} & {} & \left( 13 \right) \\\alpha =x\left( {{0}^{+}} \right)-x\left( \infty \right) & {} & {} \\\end{matrix}$

Substitute for α in **equation 12** to find the complete response:

$x\left( t \right)=\left[ x\left( {{0}^{+}} \right)-x\left( \infty \right) \right]{{e}^{-t/\tau }}+x\left( \infty \right)\begin{matrix}{} & t\ge 0 & \left( 14 \right) \\\end{matrix}$

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]]>The post Maximum Power Transfer Theorem appeared first on Electrical A2Z.

]]>Maximum power transfer theorem states that maximum power output is obtained when the load resistance R

_{o}is equal to Thevenin resistance R_{T}as seen from load Terminals.

The reduction of any **linear resistive circuit** to its **Thevenin** or Norton equivalent form is a very convenient conceptualization, as far as the computation of load-related quantities is concerned. One such computation is that of the power absorbed by the load.

The **Thevenin** and **Norton** models imply that some of the power generated by the source will necessarily be dissipated by the internal circuits within the source. Given this unavoidable power loss, a logical question to ask is, how much power can be transferred to the load from the source under the most ideal conditions? Or, alternatively, what is the value of the load resistance that will absorb maximum power from the source? The answer to these questions is contained in the **maximum power transfer theorem**, which is the subject of this section.

**Figure 1** Power transfer between source and load

The model employed in the discussion of power transfer is illustrated in **Figure 1,** where a practical source is represented by means of its Thevenin equivalent circuit. The power absorbed by the load *P _{o}* is:

\[\begin{matrix}{{P}_{o}}=i_{o}^{2}{{R}_{o}} & {} & \left( 1 \right) \\\end{matrix}\]

And the load current is:

\[\begin{matrix}{{i}_{o}}=\frac{{{v}_{T}}}{{{R}_{o}}+{{R}_{T}}} & {} & \left( 2 \right) \\\end{matrix}\]

Combining the two expressions, the load power can be computed as:

\[\begin{matrix}{{P}_{o}}=\frac{v_{T}^{2}}{{{\left( {{R}_{o}}+{{R}_{T}} \right)}^{2}}}{{R}_{o}} & {} & (3) \\\end{matrix}\]

The expression for *P _{o} *can be differentiated with respect to

\[\begin{matrix}\frac{d{{P}_{o}}}{d{{R}_{o}}}=0 & {} & \left( 4 \right) \\\end{matrix}\]

Plug in for *P _{o}* and solve to obtain:

\[\begin{matrix}\frac{d{{P}_{o}}}{d{{R}_{o}}}=\frac{v_{T}^{2}{{\left( {{R}_{o}}+{{R}_{T}} \right)}^{2}}-2v_{T}^{2}{{R}_{o}}\left( {{R}_{o}}+{{R}_{T}} \right)}{{{\left( {{R}_{0}}+{{R}_{T}} \right)}^{4}}} & {} & \left( 5 \right) \\\end{matrix}\]

Thus, at the maximum value of P_{o} the following expression must be satisfied.

\[\begin{matrix}{{\left( {{R}_{o}}+{{R}_{T}} \right)}^{2}}-2{{R}_{o}}\left( {{R}_{o}}+{{R}_{T}} \right)=0 & {} & \left( 6 \right) \\\end{matrix}\]

The solution of this equation is:

$\begin{matrix}{{R}_{o}}={{R}_{T}} & {} & \left( 7 \right) \\\end{matrix}$

So maximum power transferred is;

\[{{P}_{max}}=\frac{V_{T}^{2}}{4{{R}_{T}}}\]

We got above expression by substituting R_{T}=R_{o} into equation (3).

Thus, to transfer maximum power to a load, the equivalent source and load resistances must be matched, that is, equal to each other. **Figure 2** depicts a plot of the load power divided by $v_{T}^{2}$ versus the ratio of *R _{o} *to

**Figure 2** Graphical representation of maximum power

This analysis shows that to transfer maximum power to a load, given a fixed equivalent source resistance, the load resistance must match the equivalent source resistance. What if the problem statement were reversed such that the maximum power transfer to the source resistance is sought for a fixed load resistance? What would be the value of the source resistance that maximizes the power transfer in this case?

**Figure 3** Source loading effects

A problem related to power transfer is that of **source loading**. This phenomenon, which is illustrated in **Figure 3**, may be explained as follows: When a practical voltage source is connected to a load, the source current through the load will cause a voltage drop across the internal source resistance v_{int}; as a consequence, the voltage actually seen by the load will be somewhat lower than the open-circuit voltage of the source. As stated earlier, the *open-circuit* *voltage* is the Thevenin voltage. The extent of the internal voltage drop within the source depends on the amount of current drawn by the load. With reference to **Figure 3,** this internal drop is equal to *iR _{T}*, and therefore the load voltage will be

\[\begin{matrix}{{v}_{o}}={{v}_{T}}-i{{R}_{T}} & {} & \left( 8 \right) \\\end{matrix}\]

It should be apparent that it is desirable to have as small an internal resistance as possible in a practical voltage source.

In the case of a current source, the internal resistance will draw some current away from the load because of the presence of the internal source resistance; this current is denoted by i_{int} **in Figure 3**. Thus, the load will receive only part of the *short-circuit current* (the Norton current) available from the source:

\[\begin{matrix}{{i}_{o}}={{i}_{N}}-\frac{v}{{{R}_{T}}} & {} & \left( 9 \right) \\\end{matrix}\]

It is therefore desirable to have a very large internal resistance in a practical current source.

**Find R _{o}**

**Solution**

Let’s find V_{T} first across 150 Ω resistance

${{V}_{T}}=360*\frac{150}{150+30}$

${{V}_{T}}=300~V$

To find R_{T }, short circuit the voltage source

${{R}_{T}}=150~||~30=25\Omega $

So, for maximum power transfer, we know that

\[{{R}_{o}}={{R}_{T}}=25\text{ }\Omega\]

Now, Find Maximum power transfer to the load

${{P}_{max}}=\frac{V_{T}^{2}}{4{{R}_{T}}}=900~W$

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]]>The post Superposition Theorem appeared first on Electrical A2Z.

]]>For any linear circuit, the principle of superposition states that each independent source contributes to each voltage and current present in the circuit. Moreover, the contributions of one source are independent of those from the other sources. In this way, each voltage and each current in a circuit of N independent sources is the sum of N component voltages and N component currents, respectively. |

As a problem-solving tool, the principle of superposition permits a problem to be decomposed into two or more simpler problems. The efficiency of this “divide and conquer” tactic depends upon the particular problem being solved. However, it may enable a simple closed-form solution of an otherwise complicated symbolic circuit problem, where node and mesh analyses may offer little help.

The method is to turn off (set to zero) all independent sources but one, and then solve for voltages and currents due to the lone remaining independent source. This procedure may be repeated successively for each source until the contributions due to all the sources have been computed. The components for a particular voltage or current can be summed to find its value in the original complete circuit.

A zero-voltage source is equivalent to a **short circuit** and a zero current source is equivalent to an **open circuit**. When using the principle of superposition, it is necessary, and helpful, to replace each zero source with its equivalent short- or open-circuit and thus simplify the circuit. These substitutions are summarized in **Figure 1.**

Superposition may be applied to circuits containing ** dependent sources**; however, the dependent sources must not be set to zero. They are not independent sources and must not be treated as such. To do so, would lead to an incorrect result.

**Figure 1** Zeroing voltage and current sources

**1.** Define the voltage *V* or current *I* to be solved in the circuit.

**2.** For each of the *N* sources, define a component voltage *v*_{k} or current *i _{k}* such that

\[\begin{matrix}V={{v}_{1}}+{{v}_{2}}+…..+{{v}_{N}} & Or & I={{i}_{1}}+{{i}_{2}}+…..+{{i}_{N}} \\\end{matrix}\]

3. Turn off all sources except source* S _{j}* and solve for the component voltage

4. Find the complete solution for the voltage *V* or current *I* by summing all of the components as defined in **step 2**.

**Details and Examples**

An elementary application of the principle is to find the current in a single loop with two sources connected in series, as shown in **Figure 2.**

**Figure 2** The principle of superposition

The current in the far-left circuit of **Figure 2** is easily found by a direct application of KVL and Ohm’s law.

$\begin{matrix}{{v}_{B1}}+{{v}_{B2}}-iR=0 & {} & {} \\Or & {} & \left( 1 \right) \\i=\frac{{{v}_{B1}}+{{v}_{B2}}}{R} & {} & {} \\\end{matrix}$

**Figure 2** depicts the far-left circuit as being equivalent to the combined effects of two component circuits, each containing a single source. In each of these two circuits, one battery (which is a DC voltage source) has been set to zero and replaced with a short-circuit.

KVL and Ohm’s law can be applied directly to each of these component circuits.

\[\begin{matrix}{{i}_{B1}}=\frac{{{v}_{B1}}}{R} & and & \begin{matrix}{{i}_{B2}}=\frac{{{v}_{B2}}}{R} & \left( 2 \right) \\\end{matrix} \\\end{matrix}\]

According to the principle of superposition

\[\begin{matrix}i={{i}_{B1}}+{{i}_{B2}}=\frac{{{v}_{B1}}}{R}+\frac{{{v}_{B2}}}{R}=\frac{{{v}_{B1}}+{{v}_{B2}}}{R} & {} & \left( 3 \right) \\\end{matrix}\]

Voila! The complete solution is found, as expected. This simple example illustrates the essential method; however, more challenging examples are needed to reinforce it.

Find V in the circuit given below:

After turning off current source, we have

Using voltage divider;

${{V}_{1}}=\frac{4}{4+8}*6=2V$

Now, turning off the voltage source would give us the following result;

Using current divider;

${{i}_{3}}=\frac{8}{4+8}*3=2A$

Whereas;

${{V}_{2}}=4*{{i}_{3}}=4*2=8V$

The algebraic sum of contribution from all sources would be;

$V={{V}_{1}}+{{V}_{2}}=2+8$

$V=10V$

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