In this page, we learn more about the the frequency response of second-order circuits. Based on the response, we can classify them as listed below.
It is first shown how the frequency response of a system, given its transfer function. Later a circuit that corresponds to that transfer function is presented. We start off with a second-order low pass circuit. This page presents details about the low-pass circuit. The other circuits are presented in the subsequent pages.
The transfer function of a second-order over-damped low-pass system is presented below.
Equation (3.1) presents the the transfer function of a second-order over-damped low-pass system in terms of Laplace transform. Equation (3.2) presents the transfer function in the phasor form. The magnitude and the phase angle of the phasor can be represented as a function of frequency, as shown by equation (3.3). Since the second-order transfer function in equation (3.1) contains two distinct poles, the system is over-damped. We can obtain the polar plot from equation (3.3). This plot is presented later. Let us first see how we can obtain the asymptotic log magnitude plot.
Given the transfer function in factored form, it is easy to obtain the asymptotic magnitude. Each factor, be it of first-order or of second-order, has a real part and imaginary part. At any ω, the real part and the imaginary part of a factor can be evaluated. For getting the asymptotic magnitude, retain for each factor the greater of the two parts and ignore the other lesser part. When both the parts are equal, it does not matter which part is retained and which is ignored. For the transfer function in (3.1), assume that p1 < p2. When ω < p1 < p2, the asymptotic magnitude is obtained, as shown above. When p1 < ω < p2, the asymptotic magnitude is obtained, as shown below.
When p1 < p2 < ω, the asymptotic magnitude is obtained, as shown below.
It is necessary to check that the same asymptotic value is obtained at each of the corner frequencies. At a corner frequency, the asymptotic value can be evaluated from two expressions and both expressions should yield the same value at the corner frequency, as shown above.
The second-order system, defined by equation (3.2) has two corner frequencies, and they are p1 and p2. The asymptotic magnitude can be evaluated, as shown by equations (3.4), (3.5) and (3.6). The same set of equations can be presented, as shown below.
From equations (3.7), (3.8) and (3.9), we can obtain the asymptotic log magnitude plot, as shown below.
In Fig. 31, the x-axis is in log (ω), but usually the values expressed on the x-axis are frequencies themselves. Instead of marking the corner frequencies as log (p1) and log (p2), they are marked as p1 and p2.
A second-order transfer function is displayed by equation (3.10). For a definite set of values of K, p1 and p2, you can draw the asymptotic plot with the help of the applet provided below.
You can drag the lines and place them at the appropriate places. You can click on the Show Plot button to display the asymptotic plot. You can try out another problem by clicking on Try Another Problem button.
Let us find out next how to obtain the approximate phase angle plot for the transfer function.
Given the transfer function as shown by equation (3.11), the transfer function can be expressed as a function of ω, as shown by equation (3.12) . The phase angle can be expressed as shown by equation (3.13).
We can decompose the phase angle into two parts, as expressed by equations (3.14) and (3.15). The total phase angle is the sum of the two parts, as shown by equation (3.16).
An applet is presented below, to illustrate how the approximate phase angle plot can be obtained. The value of p1 remains fixed, whereas the value of p2 is set to be one of the three values, 5, 50 or 500. The value of p1 remains fixed at 1. When the value of p2 is 5, it is less than 10p1. When the value of p2 is 50, it is more than 10p1 but the value of 0.1 p2 is less than 10p1. When the value of p2 is 500, the value of 0.1 p2 is more than 10p1. Hence to get the approximate plot of phase angle, a lot of care is needed. You can drag the lines and place them at the appropriate locations. You can verify your solution by clicking on Show Plot button. Then the plot of the approximate phase angle is shown in red colour, whereas the plot of the actual phase angle is shown in blue colour. You may find that it is not that easy to draw the approximate phase angle plot. When the transfer function gets to be complex, it is easier to sketch the actual phase angle plot.
An RLC circuit, that can perform as a low-pass filter, is shown below.
The transfer function for the circuit is shown below.
For the specified values of components, we get the transfer function of an over-damped low-pass second-order system. It has been explained above how we can obtain the log-magnitude plot and the phase-angle plot of an over-damped system.
Next we find out how we can obtain the polar plot. The polar plot presents real part and the imaginary part, with the angular frequency as the parameter that is varied. The polar plot of equation (3.18) is shown below.
The Matlab script used to get the polar plot is shown below.
% Program: Polar Plot of Equation 3.18 hh= 10^0.05; omega = 0.01; n = 1; while (omega < 100) c = 16/(2 + j*omega)/(8 + j*omega); ReH(n) = real(c); ImH(n) = imag(c); omega = omega*hh; n = n +1; end;
plot(ReH, ImH)
title('Polar Plot of Equation 3.18')
xlabel('Real Part')
ylabel('Imaginary Part')
axis([-0.2,1,-0.7,0])
grid
A second-order low-pass system can even be critically-damped or under-damped. For the RLC circuit in Fig. 34, we get the following equations.
The polar plots can be obtained easily. The matlab scripts and the polar plots obtained for equations (3.20) and (3.22) are shown below.
% Program: Polar Plot of Equation 3.20 hh= 10^0.05; omega = 0.01; n = 1; while (omega < 100) c = 16/((4 + j*omega)*(4 + j*omega)); ReH(n) = real(c); ImH(n) = imag(c); omega = omega*hh; n = n +1; end;
plot(ReH, ImH)
title('Polar Plot of Equation 3.20')
xlabel('Real Part')
ylabel('Imaginary Part')
axis([-0.2,1,-0.7,0])
grid
% Program: Polar Plot of Equation 3.22 hh= 10^0.05; omega = 0.01; n = 1; while (omega < 100) c = 25/(25 - (omega*omega)+ j*6*omega); ReH(n) = real(c); ImH(n) = imag(c); omega = omega*hh; n = n +1; end;
plot(ReH, ImH)
title('Polar Plot of Equation 3.20')
xlabel('Real Part')
ylabel('Imaginary Part')
axis([-0.3,1,-1.0,0])
grid
When the system is under-damped, it can be seen that the real part of the transfer function becomes more negative. This aspect gets more pronounced if the damping factor reduces.
The process of obtaining the asymptotic plots is slightly different, when the system is critically or under-damped.
Given the transfer function of a critically-damped system as in equation (3.23), its complex value and the angle can be represented, as shown by equations (3.24) and (3.25).
The asymptotic plot can be obtained as shown below.
The approximate phase angle plot can be obtained as shown below.
The asymptotic log-magnitude plot and the approximate phase angle plot of a critically-damped system are shown below.
It can be asked whether the approximation adopted for getting linear phase angle plot is really the best method. An applet is presented below, wherein you can set the slope per decade and see the plot.
For each setting of the slope, the maximum deviation is estimated and it is displayed. It can be seen that the plot for the approximate phase angle has the lowest deviation from the actual plot when the slope is - 90o per decade. Next we find out how we can draw the frequency response for an under-damped system.
The transfer function of an under-damped is presented below by equation (3.32). For an under-damped system, the damping factor ξ has a value lower than one. For a passive network, shown in Fig. 34, the gain K is equal to one. In equation (3.32), ωn is the frequency at which the real part of the denominator is equal to zero.
The technique to get the asymptotic plot is shown below.
The imaginary term is ignored for the asymptotic log-magnitude plot. The real part is chosen as shown.
For the circuit in Fig. 34, let C = 0.1 F, L = 0.1 H and R=0.5 Ω. Then
For under-damped system, the slope of the asymptotic curve changes abruptly from 0 to - 2 once ω reaches the value of ωn . The value of ωn for the transfer function in equation (3.35) is 10 rad/s. The Matlab script for obtaining the Bode plot is shown below.
% Bode Plots for 1/[(1+as+bs^2)] clear a=0.05; b=0.01;
y = 10^0.025; omega = 1;
for m=1:80; theta(m) = omega; if (omega<1/sqrt(b)) asGain(m)=1.0; else asGain(m)=1.0/(omega*omega*b); end; actVal=1/(1+j*a*omega-b*omega*omega); mag(m)= actVal; phase(m)=180/pi*angle(mag(m)); omega = omega*y; end;
subplot(2,1,1)
loglog(theta,abs(mag),theta,asGain)
ylabel('magnitude')
axis([1 100 0.01 2.5])
grid on
subplot(2,1,2)
semilogx(theta,phase)
ylabel('phase angle')
xlabel('w, angular frequency')
axis([1 100 -180 0])
grid on
gtext('slope -2')
The plots obtained are shown below.
For an under-damped system, the magnitude of response tends to peak at some frequency. It is seen that the lower the damping factor is, the higher the peak is. We can obtain the peak value of frequency response and the frequency at which it occurs as follows.
We can normalize the above equation, as shown below, where ωN is the normalized frequency. Note how the normalized frequency, ωN is defined.
We can get a set of plots for various values of damping factor, ξ. The Matlab script used is presented below.
% Frequency Response of under damped transfer function % damping factor df varied from 0.1 to 0.9 clear for k=1:5; df(k)=0.2*k-0.1; for n=1:150; w(n)=(1.02^n)-1; va(n)=(1.0-w(n)*w(n))^2; vb(n)=(2*df(k)*w(n))^2; mag(k,n)=1/sqrt(va(n)+vb(n)); end; end;
for n=1:150;
v_1(n)=mag(1,n);
v_3(n)=mag(2,n);
v_5(n)=mag(3,n);
v_7(n)=mag(4,n);
v_9(n)=mag(5,n);
end;
loglog(w,v_1,w,v_3,w,v_5,w,v_7,w,v_9)
ylabel('magnitude')
xlabel('w, angular frequency')
grid on;
gtext('df=0.1')
pause
gtext('df=0.5')
pause
gtext('df=0.9')
A polar plot can be obtained for the normalized transfer function, with the damping factor varied from 01. to 0.9. The Matlab script used for obtaining the set of plots is shown below.
% Polar Response of under damped transfer function % damping factor df varied from 0.1 to 0.9 clear for k=1:5; df(k)=0.2*k-0.1; for n=1:1000; w(n)=log10((1.1)^(n-1)); va(n)=1.0-w(n)*w(n); vb(n)=2*df(k)*w(n); mag(k,n)=1/(va(n)+j*vb(n)); end; end;
for n=1:1000; v_r1(n)=real(mag(1,n)); v_i1(n)=imag(mag(1,n)); v_r3(n)=real(mag(2,n)); v_i3(n)=imag(mag(2,n)); v_r5(n)=real(mag(3,n)); v_i5(n)=imag(mag(3,n)); v_r7(n)=real(mag(4,n)); v_i7(n)=imag(mag(4,n)); v_r9(n)=real(mag(5,n)); v_i9(n)=imag(mag(5,n)); end;
plot(v_r1,v_i1,v_r3,v_i3,v_r5,v_i5,v_r7,v_i7,v_r9,v_i9)
grid on
ylabel('Imaginary Part')
xlabel('Real Part')
gtext('df=0.1')
pause
gtext('df=0.9')
pause
gtext('<--')
pause
gtext('w')
An applet is presented below to show the asymptotic and the actual log-magnitude response of a second-order under-damped system, and this applet makes use of equation (3.43), wherein the frequency is normalized. You can set the value of the damping factor, ξ and a coefficient, k. The significance of k is explained below. The applet also plots the actual phase angle and the approximate angle as a function of ω , derived from the value of k and ξ .
The approximate plot of the phase is obtained as shown below.
The best fit seems to be obtained for k = 1.
Usually the filter circuits are built using op amps. The Sallen-Key filter circuit is presented below.
We can obtain the transfer function of the circuit in Fig. 36, as shown below.
Since the source voltage equals the sum of node voltage V1(s) and the drop across Y1, we get equation (3.44). We can form a node equation at the node with voltage V1(s), and equation (3.45) is the equation formed at this node. Since the non-inverting terminal of the op-amp is grounded, the inverting terminal acts as the virtual ground.
Equation (3.46) presents an expression for current I2. By replacing V1(s) by the expression obtained from equation (3.45), we get equation (3.46). Equation (3.47) presents an expression for current I4, wherein V1(s) is replaced by the expression obtained from equation (3.45). From equations (3.45), (3.46) and (3.47), we get equation (3.48), which presents current I1 in terms of the circuit admittances and the output voltage of the op-amp. From equation (3.44), we can express current I1 in terms of the source voltage VS(s) and the output voltage VO(s).
Then source voltage VS(s) can be expressed in terms of the circuit admittances and the output voltage, as shown by equation (3.49). The transfer function of the circuit is expressed by equation (3.50). By proper choice of elements for admittances, the transfer function in equation (3.50) can represent a low-pass filter circuit.
For the circuit in Fig. 37, we can express the transfer function slightly differently, as shown below.
Equation (3.54) expresses the transfer function of an under-damped low-pass filter when the value of ξ is less than 1. We can choose the components, such that the transfer function represents an under-damped system.
A set of values to yield a low-pass filter is presented above.
% Second-order Low Pass Filter: Sallen Key Filter Circuit
clear;
Ao=10;
df=0.6;
for n=1:175;
af(n)=(1.05^n)/100;
num(n)=Ao;
den1(n)=1-(af(n)*af(n));
den2(n)=2*df*af(n);
mag(n)=num(n)/(den1(n)+j*den2(n));
phase(n)=180/pi*angle(mag(n));
end;
subplot(2,1,1)
loglog(af,abs(mag))
ylabel('Magnitude')
grid on;
subplot(2,1,2)
semilogx(af,phase)
ylabel('Phase angle')
xlabel('normalized angular frequency')
grid on;
In this page, we studied how we can get the frequency response of a second-order low-pass circuit. In the next page, we take up the topic of the frequency response of a second-order high-pass circuit.