This page is on the frequency response of first-order circuits. The frequency response of single elements is presented first, followed by frequency response of first-order circuits. Circuits such as the low-pass circuit, the high-pass circuit, and the lead-lag networks are presented. The topic of frequency response is presented from more than one angle, and this type of approach may enhance the understanding of the vital topic of frequency response.
Let us start with the frequency response of a single element that has a single zero at the origin. This system has the pole at infinity.
Given the transfer function, as stated by equation (1.1), we get equation (1.2). The magnitude and the phase angle of the transfer function in equation (1.1) are presented by equation (1.3).
The plots of the transfer function, defined by equation (1.1) are presented in Fig. 1. We can make some observations, which are listed below.
Now let us consider an inductor. The reactance of an inductor is expressed by equations (1.5) and (1.6). Equation (1.7) presents the magnitude and the angle associated with the reactance. We can express the log of the magnitude, as shown by equation (1.8).
The log-magnitude plots in Fig.2 show how the magnitude varies as a function of frequency and as a function of inductance. As the inductance varies, the plot shifts up or down.
Now let us consider a capacitor. The admittance of an inductor is expressed by equations (1.9) and (1.10). Equation (1.11) presents the magnitude and the angle associated with the admittance. We can express the log of the magnitude, as shown by equation (1.12).
The log-magnitude plots in Fig.3 show how the magnitude of admittance of a capacitor varies as a function of frequency and as a function of capacitance. As the capacitance varies, the plot shifts up or down.
Next, let us deal with the frequency response of a single element that has a single pole at the origin. This system has the zero at infinity.
Given the transfer function, as stated by equation (1.13), we get equation (1.14). The magnitude and the phase angle of the transfer function in equation (1.13) are presented by equation (1.15).
The plots of the transfer function, defined by equation (1.13) are presented in Fig. 4. We can make some additional observations to supplement those presented earlier.
Now let us consider the reactance of a capacitor.
The reactance of a capacitor is expressed by equations (1.17) and (1.18). Equation (1.19) presents the magnitude and the angle associated with the reactance. We can express the log of the magnitude, as shown by equation (1.20).
The log-magnitude plots in Fig.4 show how the magnitude of capacitive reactance varies as a function of frequency and as a function of capacitance. As the capacitance varies, the plot shifts up or down.
Before a low-pass circuit is presented, let us consider a transfer function that represents a low-pass system.
Equation (1.21) presents the transfer function of a first-order low-pass system. It is explained below why this system is called a low-pass system. The transfer function can be represented in either of the two ways, with the standard form on the left-side being the preferred form. The transfer function can be expressed as a complex value, with the magnitude and the phase function varying as a function of angular frequency, ω. Equations (1.22) and (1.23) present the transfer function as a complex value.
The frequency response of a system can be presented either as a combination of log-magnitude and phase angle plots or as a polar plot. The log-magnitude plot or the Bode plot is the preferred technique of presentation. The reasons are listed below.
Now let us see how we can obtain the asymptotic plot for the transfer function, expressed by equation (1.21).
Equation (1.22) has only one factor in the denominator, called here as df(jω). For each value of ω, evaluate df(jω). When the real part of df(jω) is greater than its imaginary part, ignore the imaginary part and evaluate |H(jω)|, as expressed by equation (1.25).
For each value of ω, evaluate df(jω). When the imaginary part of df(jω) is greater than its real part, ignore the real part and evaluate |H(jω)|, as expressed by equation (1.26).
For each value of ω, evaluate df(jω). When the imaginary part of df(jω) equals its real part, ignore either one part and evaluate |H(jω)|, as expressed by equation (1.27).
The maximum error between the actual magnitude plot and the asymptotic plot occurs at the corner frequency, where the imaginary part of df(jω) equals its real part. The maximum error can be evaluated, as shown by equations (1.28), (1.29) and (1.30). It equals -0.15 or -3 dB.
Now let us consider the transfer function, specified below by equation (1.31).
The log-magnitude plots obtained are displayed below.
The phase angle plots obtained are shown below.
The Matlab script used to produce the above plots is shown above. Some comments have been added to the plots obtained.
% Program To illustrate First-order Asymptotic Plot
hh= 0.01;
for n=1:201;
val = (n-1)*hh -1;
omega = 10^val;
t(n)=omega;
if (n<101) Hasym(n) = 0;
else Hasym(n) = 1/omega;
end;
den = sqrt(1+omega*omega);
Hact(n) = 1/den;
phAct(n)= -atan(omega)*180/pi;
val = (n-1)*hh;
phAsym(n) = - val*45.0;
end;
loglog(t,Hasym,'r-',t,Hact,'b-')
title('Asymptotic and Actual Log Plots')
xlabel('Normalized Angular Frequency')
ylabel('Magnitude')
grid
pause
semilogx(t,phAsym, 'r-',t,phAct,'b-')
axis([0 10 -90 0])
title('Phase Angle: Approximate & Actual')
xlabel('Normalized Angular Frequency')
ylabel('Degrees')
grid
It is assumed that you can sketch the actual log-magnitude plot and the actual phase angle plot. For various values of ω, calculate the magnitude and the phase angle from equation (1.31). Plot the log of the magnitude with respect to the log of the frequency. Plot the phase angle with respect to the log of the frequency. The question now is how to plot the asymptotic log-magnitude plot and the approximate phase angle plot. The procedure is illustrated now.
Equation (1.32) presents the transfer function of a low-pass first-order system. It can be represented by two equations, based on the value of ω. From equations (1.33) and (1.34), we can form equations for the two lines that form the asymptotic plot.
Equation (1.35) is the equation of the flat line and equation (1.36) is the equation of the line with a slope of - 1. The sketch in Fig. 8 illustrates how the asymptotic plot is to be drawn.
The frequency range from 0 < ω < (1/τ) is known as the pass band and the frequency range above (1/τ) is known as the stop band. What it means is that the most of input signal below the frequency (1/τ) is passed to the load connected across the capacitor, and that the most of input signal above the frequency (1/τ) is shunted away from the load connected across the capacitor.
The approximate plot of the phase angle can be drawn as shown below.
The transfer function expressed by equation (1.32) defines a low-pass system. The output of the system is relatively high at low frequencies, below the corner frequency, also called as the cut-off frequency. When the frequency increases beyond the cut--off frequency, the output falls. Since the system passes low-frequency signals to the output and acts as a shunt or bypass at high frequencies, the transfer function expressed by equation (1.32) is that of a low-pass system.
Now a few low-pass circuits are presented.
The transfer function for the circuit in Fig. 10 is presented below.
Another low-pass RC circuit is presented below.
The transfer function for the circuit in Fig. 11 is presented below.
It is possible to have a low-pass RL circuit also.
The transfer function for the circuit in Fig. 12 is presented below.
POLAR PLOT
We can obtain the polar plot for the low-pass circuit. Refer to Fig. 10. Let RC = 1. Then the polar plot is obtained, as presented below.
The polar plot presents the imaginary part and the real part as a function of ω. The trajectory or the locus of the magnitude of output is along the circumference of a semi-circle. For the circuit in Fig. 10, the output is the voltage across the capacitor. The phasor sum of the voltage across the capacitor and the voltage across the resistor equals the source voltage.
The voltage across the resistor is in phase with its current whereas this current leads the capacitor voltage by 90o. For a triangle with the diameter of a semicircle as its base, the angle at the vertex of a triangle located on the semicircle is 90o. Since the angle between the voltage across the capacitor and the voltage across the resistor is 90o, the angle subtended at the periphery of the locus is 90o, and hence the locus is along a semi-circle.
For the circuit in Fig. 10, the values of the transfer function H(jω) at the two limits are as presented by equation (1.43). The polar plot of H(jω) is shown in Fig. 13. The polar plot provides information about how the magnitude and the phase angle vary as a function of ω. It is seen that the amplitude of response decreases as the angular frequency, ω increases. This means that |H(jω)| is significantly higher for lower values of ω than its value for higher values of ω . In other words, the network passes low frequency signals to the load that may be connected to the output terminals and shunts or bypasses the high frequency signals away from the load. Hence this network is called the low-pass network.
The output of the circuit in Fig. 10 lags the input voltage by angle θ, where
Hence this circuit is known as a lag network.
This circuit has another name too. At times, this circuit is known as the integrating circuit. To explain why this circuit is called so, it is preferable to use the time response of the circuit.
Let the input to the circuit be a unit step function, as specified by equation (1.45). Then the unit-step response of this circuit can be obtained, and it is expressed by by equation (1.46). On expanding the exponential term into a series, we get equation (1.47). When t << τ, we get equation (1.48). It is seen from equation (1.48) that for values of t very much less than τ, the low-pass network behaves like an integrating network.
Before a high-pass circuit is presented, let us consider a transfer function that represents a high-pass system.
Equation (2.1) presents the transfer function of a first-order high-pass system. This system has a zero at the origin, and a pole at ω = 1/τ. It can be represented in terms of angular frequency, ω, as shown by equation (2.2). Equation (2.2) presents the transfer function both in the cartesian form, as well as in the polar form.
We can draw the asymptotic plot from equations (2.3), (2.4) and (2.5). Equation (2.3) presents the log-magnitude of the asymptotic plot as a function of ω, where the value of ω lies in the range specified. When ωτ is equal or above unity, the log of the magnitude is obtained as expressed by equations (2.4) and (2.5).
The maximum error in the log magnitude plot when the value of ω equals the corner frequency or the frequency corresponding to the pole of the system. The maximum error is 3 dB. The corner frequency is also known as the 3 dB frequency. For the transfer function in equation (2.1) let k = 1 and τ = 1. For this system, the asymptotic plot obtained is presented below.
The approximate plot of the phase angle as a function of log10 ω is obtained as shown below.
When ωτ < 0.1, the imaginary part of the denominator is ignored and the phase angle is then equal to 90o, due to zero at the origin. In the range specified by 0.1 < ωτ < 10.0, the phase angle falls linearly with respect to log10 ω, and it remains at 0o for ωτ > 10.0. For the transfer function in equation (2.1) let k = 1 and τ = 1. For this system, the approximate phase angle plot obtained is presented below.
The Matlab script used to produce the above plots is shown above. Some comments have been added to the plots obtained.
% Program To illustrate First-order Asymptotic Plot
% First-order High Pass circuit
hh= 0.01;
for n=1:201;
val = (n-1)*hh -1;
omega = 10^val;
t(n)=omega;
if (n<101) Hasym(n) = omega;
else Hasym(n) = 1;
end;
den = sqrt(1+omega*omega);
Hact(n) = omega/den;
phAct(n)= 90 - atan(omega)*180/pi;
val = (n-1)*hh;
phAsym(n) = pi/2 - val*45.0;
end;
loglog(t,Hasym,'r-',t,Hact,'b-')
title('Asymptotic and Actual Log Plots')
xlabel('Normalized Angular Frequency')
ylabel('Magnitude')
grid
pause
semilogx(t,phAsym, 'r-',t,phAct,'b-')
axis([0 10 0 90])
title('Phase Angle: Approximate & Actual')
xlabel('Normalized Angular Frequency')
ylabel('Degrees')
grid
The asymptotic plot for the log-magnitude and the approximate plot for the phase angle have been presented in Figs. 14 and 15. It is shown next how they can be obtained and plotted by hand.
The sketch in Fig. 16 shows how the asymptotic plot is to be drawn, for the transfer function displayed as part of Fig. 16. First locate the point of intersection. Draw a line with a slope of 1 starting from this point covering the frequency range below 1/τ. This line with a slope of 1 represents equation (2.3). The point of intersection is represented by equation (2.4). The horizontal line, drawn from the point of intersection covering frequency range above 1/τ, represents equation (2.5).
The approximate plot of the phase angle is presented above. This plot is derived from equations (2.8), (2.9) and (2.10).
Based on discussion on the low-pass and the high-pass circuit, some properties of the frequency response can be defined.
At each pole frequency, the slope of asymptotic gain plot changes by - 1. For example, if the slope of this plot at a frequency below the pole frequency is 0 with respect to log ω, the slope of the asymptotic plot changes to -1 or - 20 dB/decade with respect to log ω at the pole frequency, as shown by the plot displayed above.
At each zero frequency, the slope of asymptotic gain plot changes by + 1. For example, if the slope of this plot at a frequency below the zero frequency is - 1, or -20 dB/decade, the slope of the asymptotic plot becomes zero at the zero frequency.
For variation of ω from zero to infinity, the change in phase angle due to each pole is -90o and the change due to each zero is 90o. At the pole frequency, the contribution to phase angle of H(jω ) due to that pole is -45o and at the zero frequency, the contribution to phase angle of H(jω ) due to that zero is 45o. At the break points, the real part equals the imaginary part due to the first-order factor containing the pole or zero..
Usually the asymptotic gain is evaluated for ω = 1. It is explained later how the gain can be evaluated, with the help of some examples.
Even though the gain and frequency are plotted on a log scale, the values indicated in the plots are the actual values. This method enables one to read off the response at any ω directly, without recourse to antilogarithms.
The range of frequencies over which the gain is significantly attenuated is known as the stop band. For a low-pass filter, the stop band starts at the corner frequency and the range is defined by ( 1/τ < ω < ∞ ) is the stop band . For the high-pass filter, the defined by (0 < ω < 1/τ) is the stop band.
The range of frequencies over which there is little attenuation in gain is known as the pass band. For a low-pass filter with the corner frequency at ω = 1/ τ , the range defined by (0 < ω < 1/τ) is the pass band. For a high-pass filter with the corner frequency at ω = 1/ τ , the range defined by ( 1/τ < ω < ∞ ) is the pass band.
The frequency associated with the boundary that separates the stop band and the pass band is called the cut-off frequency. For a first-order type-zero system, the corner frequency is also the cut-off frequency.
At cut-off frequency, the gain is 0.707 times the gain in pass band. The square of the gain represents power and it is half of that in the pass band. Hence the cutoff frequency is also called the half-power frequency.
The bandwidth of a gain characteristic is defined as the frequency range of its pass band.
Bode plots are used when the response over a wide frequency range is to be displayed. If the frequency response over a range is to be displayed, the Bode plots are used if the ratio of is large, say higher than 100.
Next a high-pass RC circuit is shown.
At the output of the RC circuit, an op amp buffer has been provided to eliminate the loading effect. Such a buffer is required for low pass circuits, displayed in Figs. 10, 11 and 12 also. For the circuit in Fig. 18, we get the transfer function as shown below.
The circuit in Fig. 18 has unity feedback. It can be changed and the gain can be greater than one, for the circuit shown in Fig. 19.
For the circuit in Fig. 19, we get the transfer function shown below.
It is possible to have a high-pass using R and L, as shown below.
For the circuit in Fig. 20, we get the transfer function shown below.
Polar Plot
The polar plot of the equation displayed in Fig. 21 is a semicircle of unity diameter, since the angle subtended by the vertex on the circumference is a right-angle. The polar plot is shown in Fig. 21. For the RC circuit shown in Fig. 18, the source current leads the source voltage. Since the output voltage is the resistor voltage, the output voltage leads the source voltage. It is known that the voltage and the current of a resistor are in phase.
Since the output voltage leads the source voltage, this network is known as the lead network. Given that the input voltage is a step voltage, we have that
The unit-step response of the first-order high-pass filter circuit is shown in Fig. 22. If the time constant τ of the high-pass network is quite small, the time response of the high-pass network appears as the derivative of the source voltage. If sτ <<1, the transfer function in equation (2.14) can be approximated to be:
Since multiplication by s in s-domain corresponds to differentiation in time domain, it can be seen from equation (2.17) that the high-pass network can be said to be a differentiating network when sτ <<1.
A lead compensator is a circuit that is used as part of a control circuit for adjusting the phase margin. The frequency response of this circuit is presented here.
The transfer function for this network can be obtained as follows:
As a >1, the phase angle of denominator is greater than that of the numerator for any ω. Hence the output lags the input at any ω. The polar plot of H(jω) is shown in Fig. 24.
From equation (2.19), we can obtain the following equations.
For values of ω between 0 and ∞, the locus of H(jω) is a semicircle, and its diameter and the centre are stated by equations (2.21) and (2.22). It is easy to prove that the following equation.
Equation (2.23c) is an equation of a circle.
The phase angle φ of the transfer function is obtained as shown below.
Equation (2.23) specifies the phase angle of the numerator term and the denominator term of the transfer function, expressed by equation (2.18). The phase angle of H(jω) is obtained by subtracting the phase angle of the denominator term from he phase angle of the numerator term, as shown by equation (2.25). We can obtain the minimum or the most negative value of phase angle of H(jω) as shown by equation (2.25). The minimum value of the phase angle is obtained by differentiating the expression in equation (2.24) with respect to ω and estimating φ at the value of ω for which the derivative is zero. The minimum phase angle occurs when the frequency is the geometric mean of the pole and zero frequencies. Equation (2.26) expresses the minimum phase angle of the lag compensator circuit in Fig. 23.
When the phase angle has the minimum value, the phasor H(jω) is a tangent to the locus of the response, as shown by Fig 25.
The transfer function in equation (2.24) can be expressed as shown below.
Equation (2.27) defines a transfer function that has a pole p at - 1/(aτ) and a zero z at - 1/(τ), and the corner frequency due to the pole at ω = 1/(aτ) is lower than the corner frequency due to the zero at ω = 1/(τ) . For this transfer function, the sketch in Fig. 26 shows how the phase angle is calculated.
At any ω , draw a line connecting - p to ω. Similarly draw a line connecting -z to ω. Measure the angles and the phase angle is obtained as shown in Fig. 26.
To get the asymptotic plot of |H(jω)|, |H(jω)| can be approximated as follows.
When the value of ω lies between its pole and its zero , we get the asymptotic response as follows.
When the value of ω > z, we get the asymptotic response as follows.
Equations (2.30), (2.32) and (2.34) can be represented as shown.
It is difficult to get the approximate phase angle plot. Let us consider the following transfer function.
The Matlab script used to obtain the plots is shown below.
% Lead Compensator % Bode Plots of Transfer function (s+tau)/(s+a.tau) tau=1; % Time constant = 1 sec a=4; % Coefficient used to specify the location of pole p=1/(a*tau); % Pole location z=1/tau; % Zero Loaction mw=sqrt(p*z); % Geometric mean of pole and zero m = log(a)/30; % 30 geometric steps from pole and zero minphAngleRad=- atan(0.5*(sqrt(a)-1/sqrt(a))); minphAngleDeg= 180/pi*minphAngleRad; omegaMin = 1/(tau*sqrt(a))
k = p/(10^(m*30));
for n=1:90;
w(n) = k*(10^(m*n));
if (w(n)<=p) asGain(n)=1.0;
elseif (w(n)<=z) asGain(n)=1/(a*tau*w(n));
else asGain(n)=1/a;
end;
mag(n)=sqrt(1+w(n)*w(n)*tau*tau)/sqrt(1+w(n)*w(n)*tau*tau*a*a);
tval1=atan(tau*w(n));
tval2=atan(a*tau*w(n));
phase(n)=180.0/pi*(tval1-tval2);
end;
minphAngleDeg
subplot(2,1,1)
loglog(w,mag,w,asGain)
ylabel('magnitude')
grid on;
subplot(2,1,2)
semilogx(w,phase)
ylabel('phase angle')
xlabel('w, angular frequency')
grid on;
gtext('wm')
pause
gtext('min phase angle')
pause
gtext('p')
pause
gtext('z')
Fig. 27: Log-magnitude and phase angle plots
It can be seen that the asymptotic log-magnitude can be obtained from equations (2.35), (2.36) and (2.37).
Another compensating network, a lag compensator, is presented next. A lead compensator is shown in Fig. 28. The transfer function for this network can be obtained as follows:
As a < 1, the phase angle of denominator is less than that of the numerator for any ω. Hence the output leads the input at any ω. From the transfer function defined by equation (2.40), we get the following equations.
The polar plot of H(j ω) is shown in Fig. 29.
The locus of H(j ω) is a semi-circle. The phase angle φ of the transfer function defined by equation (2.40) is:
The maximum value of the phase angle is obtained by differentiating the above expression with respect to ω and estimating at a value of ω for which the derivative is zero. Then
At any ω, the phase angle can be obtained, as shown in sketch in Fig. 31.
The transfer function in equation (2.40) can be expressed as:
Equation (2.46) defines a transfer function that has a pole p at - 1/(aτ) and a zero z at - 1/(τ), and the corner frequency due to the pole at ω = 1/(aτ) is higher than the corner frequency due to the zero at t ω = 1/(τ) since a < 1. For this transfer function, the plot shown in Fig. 30 shows how the phase angle is calculated and when the maximum phase angle occurs.
From equation (2.46), we get the following equations.
We can obtain the asymptotic plot from equations (2.53), (2.54) and (2.55). At the zero frequency, the slope of the log magnitude changes from 0 to +1 and the slope changes from +1 to 0 at the pole frequency. . The Bode plots are shown below, for the case where τ =1, a = 0.25. The relevant transfer function is shown below.
The Matlab script used to obtain the Bode plots is shown below.
% Bode Plots of Transfer function a(s+tau)/(s+a.tau)
% Bode Plot for the lead compensating network
tau=1;
a=0.25;
p=1/(a*tau);
z=1/tau;
mw=sqrt(p*z);
m = log(1/a)/30; % 30 geometric steps from pole and zero maxphAngleRad= atan(0.5*(sqrt(1/a)- sqrt(a))); maxphAngleDeg= 180/pi*maxphAngleRad; omegaMax = 1/(tau*sqrt(a))
k = z/(10^(m*30)); for n=1:90; w(n) = k*(10^(m*n)); if (w(n)<=z) asGain(n)= a; elseif (w(n)<=p) asGain(n)=(w(n)/p); else asGain(n)=1; end; mag(n)=a*sqrt(1+w(n)*w(n)*tau*tau)/sqrt(1+w(n)*w(n)*tau*tau*a*a); tval1=atan(tau*w(n)); tval2=atan(a*tau*w(n)); phase(n)=180.0/pi*(tval1-tval2); end;
maxphAngleDeg
subplot(2,1,1)
loglog(w,mag,w,asGain)
ylabel('magnitude')
grid on;
subplot(2,1,2)
semilogx(w,phase)
ylabel('phase angle')
xlabel('w, angular frequency')
grid on;
gtext('wm')
pause
gtext('max phase angle')
pause
gtext('p')
pause
gtext('z')
Fig. 32: Log-magnitude and phase angle plots
This page has illustrated how we can obtain the frequency response of first-order networks. The next page shows how we can obtain the frequency response of higher-order networks.