The op amp used for the non-inverting amplifier shown in Fig. 54 has an open-loop gain with a single pole. It is defined as:
Obtain the transfer function for the non-inverting amplifier and sketch its frequency response.
SOLUTION:
The gain for the non-inverting amplifier has been obtained as
After substituting the expression for A(jω), the above equation becomes:
The frequency response of the non-inverting amplifier is presented below.
It can be seen that the band-width of the op amp with feedback is much greater than the band-width of the open-loop gain of the op amp. On the other hand, the open-loop gain of the op amp is much greater than gain of the non-inverting amplifier circuit .The increase in band-width is obtained due to the reduction in gain of the non-inverting amplifier circuit. It can also be seen that the gain-bandwidth product has the same value for the open-loop transfer function and the transfer function of the non-inverting amplifier.
Draw the log-magnitude plot for
SOLUTION:
The given transfer function can be expressed as follows, for obtaining the asymptotic response.
At first, the point (log K, 0) is marked. To represent the term K/ω, a line with a slope of -1 is drawn such that it passes through (log K,0). As T1 > 1, (1/T1) < 1 and the point (log K + log T1 , -log T1 ) lying on the line drawn is marked. The point (log K +log T1 , -log T1 ) is a break point because of the zero at (1/T1 ) . Since the line has a slope of -1 for frequencies < (1/T1 ), the log-magnitude plot has zero slope from (1/T1 <ω< 1/T2 ) . Because of the pole at - 1/T2 , the slope becomes -1 for the range (1/T2 <ω< 1/T3 ). At ω = 1/T3 , the slope becomes 0 and it changes to - 1 at ω =1/T4 . The log-magnitude plot is shown in Fig. 56.
For the transfer function expressed by equation (5.8), draw the polar and the Bode plots.
SOLUTION:
The plots are shown in Fig. 57.
For the circuit in Fig. 58, R = 10.1 Ω, L = 0.1 H, and C = 0.1 F. Draw the Bode plots for this circuit.
SOLUTION:
For the network, the transfer function is:
Then for the asymptotic plot, we can form the equations shown below.
The equation for the phase angle is shown below.
The log-magnitude plots shown below have been obtained using a Matlab program.
% Bode Plots for (s^2+100)/[(1+101s+100s^2)]
% Example 4 : Notch Filter
clear
for n=1:200;
w(n)=(1.04)^n-1.03;
if (w(n)<1) asGain(n)=1;
elseif (w(n)<10) asGain(n)=1/w(n);
elseif (w(n)<100) asGain(n)=w(n)/100;
else asGain(n)=1;
end;
mag(n)=(100-w(n)*w(n))/((1+j*w(n))*(100+j*w(n)));
phase(n)=180/pi*angle(mag(n));
end;
subplot(2,1,1)
loglog(w,abs(mag),w,asGain)
ylabel('magnitude')
grid on;
subplot(2,1,2)
semilogx(w,phase)
ylabel('phase angle')
xlabel('w, angular frequency')
grid on;
It is seen that the circuit acts as a notch filter. It is also called a band stop circuit, since it attenuates the output signal heavily between the two cutoff frequencies. For this circuit, the cutoff frequencies are 1 and 100 rad/s. It is interesting to note that the phase angle undergoes a sudden change by 180o at ω =10 rad/s since the numerator has a double zero at this frequency.
If the numerator had a complex conjugate pair of zeros instead of pair of zeros on the imaginary axis, the phase angle would not undergo a sudden change of 180o at ω =10 rad/s.
For the transfer function expressed by equation (5.8), draw the polar and the Bode plots.
SOLUTION:
The Matlab script used to obtain the plots is presented below.
% Bode Plots for Example 5
% H(s) = Num(s)/Den(s), N(s) = 100*(s + 10), Den(s) = (s + 1)(s+100)
clear
y = 10^0.025;
omega = 0.1;
for m=1:160;
theta(m) = omega;
s = j*omega;
if (omega < 1) asGain(m)=10;
elseif (omega < 10) asGain(m)= 10/omega;
elseif (omega < 100) asGain(m)= 1;
else asGain(m) = 100/omega;
end;
actVal= 100*(s + 10)/(s+1)/(s+100);
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([0.1 1000 0.1 10])
grid on
subplot(2,1,2)
semilogx(theta,phase)
ylabel('phase angle')
xlabel('w, angular frequency')
axis([0.1 1000 -90 0])
grid on
For the transfer function expressed by equation (5.20), draw the polar and the Bode plots.
SOLUTION:
Under damped System: For calculating the asymptotic magnitude, the imaginary part is ignored. For ω < 1, 1 > ω2. In this case, ω2 in denominator is ignored. For ω > 1, 1 < ω2. In this case, 1 in denominator is ignored. The phase angle decreases from 0o to - 45o, as ω increases from a low value to 1 rad/s. At ω = 1 rad/s, the phase angle is - 45o. For ω > 1, the phase angle decreases from -45o to -90o. When ω tends to ∞, the phase angle reaches -90o..
The Matlab script used to get the plots is presented below.
% Bode Plots for Example 6
% H(s) = Num(s)/Den(s), N(s) = 100*(s + 1), Den(s) = (s^2 + s + 1)
clear
y = 10^0.01;
omega = 0.1;
for m=1:200;
theta(m) = omega;
s = j*omega;
if (omega < 1) asGain(m)=100;
else asGain(m) = 100/omega;
end;
actVal= 100*(s + 1)/(s*s + s + 1);
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([0.1 10 10 200])
grid on
subplot(2,1,2)
semilogx(theta,phase)
ylabel('phase angle')
xlabel('w, angular frequency')
axis([0.1 10 -90 0])
grid on
For the transfer function expressed by equation (5.24), draw the polar and the Bode plots.
SOLUTION:
The Matlab script used to get the plots is presented below.
% Bode Plots for Example 7
% H(s) = Num(s)/Den(s), N(s) = 1 + (s/2) + (s*s)/4, Den(s) = s(1 + s/2)(1+ s/4)
clear
y = 10^0.01;
omega = 0.1;
for m=1:250;
theta(m) = omega;
s = j*omega;
if (omega < 2) asGain(m)=1/omega;
elseif (omega < 4) asGain(m)= 0.5;
else asGain(m) = 2/omega;
end;
actVal= (1 + 0.5*s + 0.25*s*s)/s/(0.5*s+1)/(0.25*s+1);
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')
grid on
axis([0.1 30 0.05 10])
subplot(2,1,2)
semilogx(theta,phase)
ylabel('phase angle')
xlabel('w, angular frequency')
grid on
axis([0.1 30 -100 -50])
Some worked examples have been presented in this page to illustrate how the frequency response of a system can be obtained. The topic of Convolution Integrals is presented next.