In this page, we learn more about the the frequency response of second-order high-pass filter circuit.
It is first shown how the frequency response of a system can be obtained , given its transfer function. Later a circuit that corresponds to that transfer function is presented. We start off with the transfer function of a second-order high-pass .
The transfer function of a second-order over-damped high-pass system is presented below.
Equation (3.54) presents the Laplace transform function of a second-order high-pass system. Equation (3.55) presents this 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 (3.56). Since the second-order transfer function in equation (3.54) contains two distinct poles, the system is over-dam aped. We can obtain the polar plot from equation (3.56). This plot is presented later. Let us see first 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 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 part. When both the parts are equal, it does not matter which part is retained and which is ignored. For the transfer function in (3.54), assume that p1 < p2. When ω > p2 < p1, the asymptotic magnitude is obtained, as shown above. When p1 < ω < p2, the asymptotic magnitude is obtained, as shown below.
It is necessary to verify that the approximation by cross-checking the value of |HAsym(jω)| at the boundary values of ω . When ω < p1< p2 , 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. The second-order system, defined by equation (3.54) has two corner frequencies, and they are p1 and p2.
The asymptotic magnitude can be evaluated, as shown by equations (3.57), (3.58) and (3.59). The same set of equations can be presented, as shown below.
From equations (3.60), (3.61) and (3.62), we can obtain the asymptotic log magnitude plot, as shown below.
In Fig. 38, 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 (3.63). 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.
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 (3.64), the transfer function can be expressed as a function of ω, as shown by equation (3.65) . The phase angle can be expressed as shown by equation (3.66).
We can decompose the phase angle into two parts, as expressed by equations (3.68) and (3.69). The total phase angle is the sum of the two parts and 180o, as shown by equation (3.69). The numerator term in equation (.365) contributes 180o to the phase angle.
An applet is presented below, to illustrate how the approximate phase angle plot can be obtained. You can change the value of p2, whereas the value of p1 remains fixed. The value of p2 is set to be one of the three values, 5, 50 or 500. 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 above circuit is obtained as shown below.
For the specified values of components, we get the transfer function of an over-damped high-pass second-order system. 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 (3.72) and the Matlab script used to get the plot are shown below.
% Program: Polar Plot of Equation 3.72
% High-pass System
hh= 10^0.05;
omega = 0.01;
n = 1;
while (omega < 100)
c = -omega*omega/(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('Fig. 40: Polar Plot of Equation 3.72')
xlabel('Real Part')
ylabel('Imaginary Part')
axis([-0.2,1,0.0,0.6])
grid
The circuit in Fig. 39 is critically-damped for the values of components specified below.
The polar plot of equation (3.74) and the Matlab script used to get the plot are shown below. The script is more or less the same as that presented for equation (3.72), except that the coefficients for ω are suitably modified.
% Program: Polar Plot of Equation 3.74
% Critically-damped High-pass System
hh= 10^0.05;
omega = 0.01;
n = 1;
while (omega < 100)
c = -omega*omega/((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('Fig. 41: Polar Plot of Equation 3.74')
xlabel('Real Part')
ylabel('Imaginary Part')
axis([-0.2,1,0.0,0.7])
grid
The circuit in Fig. 39 is under-damped for the values of components specified below.
The polar plot of equation (3.76) and the Matlab script used to get the plot are shown below. 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.
% Program: Polar Plot of Equation 3.76 % An under-damped highpass system hh= 10^0.05; omega = 0.01; n = 1; while (omega < 100) c = -omega*omega/(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('Fig. 42: Polar Plot of Equation 3.76')
xlabel('Real Part')
ylabel('Imaginary Part')
axis([-0.3,1,-.0,1.0])
grid
Let us find out how we can obtain the asymptotic plot of a critically system.
Given the transfer function of a critically-damped system as in equation (3.77), its complex value and the angle can be represented, as shown by equations (3.78) and (3.79). 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.
Next, let us find out how we can obtain the asymptotic plot of an under-damped system.
The transfer function of an under-damped is presented below by equation (3.87). For an under-damped system, the damping factor ? has a value lower than one. For the network shown in Fig. 39, the gain K is equal to one.
The technique to get the asymptotic plot is shown below.
For the circuit in Fig. 39, 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 2 to 0 once ω reaches the value of ωn . The value of ωn for the transfer function in equation (3.90) is 10 rad/s. The Matlab script for obtaining the Bode plot is shown below.
% Bode Plots for s^2/[(b+as+s^2)] % High-pass Second-order system: Equation (3.90) clear a=5; b=100;
y = 10^0.025; omega = 1;
for m=1:80; theta(m) = omega; if (omega>sqrt(b)) asGain(m)=1.0; else asGain(m)=(omega*omega/b); end; actVal=-omega*omega/(b+j*a*omega- 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 0 180])
grid on
gtext('slope 2')
The plots obtained are shown below.
Fig. 44: Frequency Response Plots: High-pass System
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 shown earlier for the low-pass system. We get the peak frequency ωm and the peak response Mm, shown below.
We can normalize the above equation, as shown below, where ?N is the normalized frequency.
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
% High-pass System Eqn. (3.96)
clear
for k=1:5;
df(k)=0.2*k-0.1;
for n=1:150;
w(n)=(1.02^n)-1;
nr(n) = w(n)*w(n);
va(n)=(1.0-w(n)*w(n))^2;
vb(n)=(2*df(k)*w(n))^2;
mag(k,n)=nr(n)/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')
Fig. 45: Bode Plots of Under-damped High-Pass System
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
% High-pass System: Eqn. (3.96)
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); nr(n) = - w(n)*w(n);
mag(k,n)=nr(n)/(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')
Fig. 46: Polar Plots of Under-damped High-Pass System
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 (3.96), wherein the frequency is normalized. You can set the value of the damping factor, ξ and a coefficient, k. The applet plots also the actual phase angle and the approximate angle as a function of ω, derived from the value of k and ξ .
The Sallen Key filter circuit has been presented before in the previous page. It is again presented below.
The transfer function obtained earlier is presented again.
We can configure the circuit in Fig. 36 as a high-pass filter circuit, as shown below.
Choose the components, as shown above. Then we obtain the transfer function as shown below, by substituting the selected admittances into equation (3.50).
The above transfer function represents a second-order high-pass filter. Let
By proper choice of component values, the filter circuit can be under-damped, critically-damped or over-damped.
The Matlab script for obtaining the plots is presented below.
% Second-order High Pass Filter: Sallen Key Circuit
clear;
Ao=10;
df=0.6;
for n=1:175;
af(n)=(1.05^n)/100;
num(n)=Ao*(af(n)*af(n));
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;
This page has explained how the frequency response of second-order high-pass filter circuit can be obtained. We look at some more second-order circuits in the next page.