Question 23.22: Consider the circuit of Figure 23.27. Using Kirchhoffs volta......
Consider the circuit of Figure 23.27. Using Kirchhoffs voltage law and Ohm’s law we obtain
v_{\mathrm{i}}=i R+v_{\mathrm{o}}For the capacitor,
v_{\mathrm{o}}=\frac{i}{\mathrm{j} \omega C}Eliminating i yields
\begin{aligned} v_{\mathrm{i}} & =v_{\mathrm{o}} \mathrm{j} \omega C R+v_{\mathrm{o}}=v_{\mathrm{o}}(1+\mathrm{j} \omega R C) \\ \frac{v_{\mathrm{o}}}{v_{\mathrm{i}}} & =\frac{1}{1+\mathrm{j} \omega R C}\quad (23.10) \end{aligned}Equation (23.10) relates the output of the system to the input of the system. Therefore,
G(\mathrm{j} \omega)=\frac{1}{1+\mathrm{j} \omega R C}It is convenient to convert G(jΩ) into polar form:
\begin{aligned} G(\mathrm{j} \omega) & =\frac{1 \angle 0}{\sqrt{1+(\omega R C)^2} \angle \tan ^{-1} \omega R C} \\ & =\frac{1}{\sqrt{1+(\omega R C)^2}} \angle-\tan ^{-1} \omega R C \end{aligned}Therefore,
|G(\mathrm{j} \omega)|=\frac{1}{\sqrt{1+(\omega R C)^2}} (23.11)
\angle G(\mathrm{j} \omega)=-\tan ^{-1} \omega R C (23.12)
The amplitude and phase characteristics for the circuit of Figure 23.27 are shown in Figure 23.28. These show the variation of |G(\mathrm{j} \omega)| \text { and } \angle G(\mathrm{j} \omega) with angular frequency \omega.
Note that the circuit is a low-pass filter; it allows low frequencies to pass easily and rejects high frequencies. The cut-off point of the filter, that is the point at which significant frequency attenuation begins to occur, can be varied by changing the values of R and C. The quantity RC is usually known as the time constant for the system. Consider the case when RC = 0.3. Equations (23.11) and (23.12) reduce to
|G(\mathrm{j} \omega)|=\frac{1}{\sqrt{1+0.09 \omega^2}} (23.13)
\angle G(\mathrm{j} \omega)=-\tan ^{-1} 0.3 \omega (23.14)
Let us examine the response of this system to a square wave input with fundamental angular frequency 1 and amplitude 1. This waveform is shown in Figure 23.29(a). We note that T = 2π. The waveform function is odd and so will not contain any cosine Fourier components. It has an average value of 0 and so will not have a zero frequency component; that is, there will be no d.c. component. Therefore calculating the Fourier components reduces to evaluating
f(t)=\sum_{n=1}^{\infty} b_n \sin \frac{2 \pi n t}{T}.
b_n=\frac{2}{T} \int_{-T / 2}^{T / 2} f(t) \sin \frac{2 \pi n t}{T} \mathrm{~d} t \quad \text { for positive integers } nSince T = 2π, we find
\begin{aligned} b_n & =\frac{1}{\pi}\left(\int_{-\pi}^0-1 \sin n t \mathrm{~d} t+\int_0^\pi \sin n t \mathrm{~d} t\right) \\ & =\frac{1}{\pi}\left(\left[\frac{\cos n t}{n}\right]_{-\pi}^0+\left[\frac{-\cos n t}{n}\right]_0^\pi\right) \\ & =\frac{1}{n \pi}(\cos 0-\cos \pi n-\cos \pi n+\cos 0) \\ & =\frac{1}{n \pi}(2-2 \cos \pi n) \\ & =\frac{2}{n \pi}(1-\cos \pi n) \end{aligned}The values of the first few coefficients are
\begin{matrix} b_1=\frac{2}{\pi }(1-cos\pi )=\frac{4}{\pi } \\b_2=\frac{2}{2\pi }(1-cos2\pi )=0 \\ b_3=\frac{2}{3\pi }(1-cos3\pi )=\frac{4}{3\pi } \end{matrix}The next stage is to evaluate the gain and phase changes of the Fourier components. Using Equations (23.13) and (23.14):
\begin{aligned} & n=1 \\ & \omega_1=1 \\ & \left|G\left(\mathrm{j} \omega_1\right)\right|=\frac{1}{\sqrt{1+0.09 \times 1}}=0.96 \\ & \angle G\left(j \omega_1\right)=-\tan ^{-1} 0.3=-16.7^{\circ} \\ & \end{aligned}.
\begin{aligned} &n=3\\ &\begin{aligned} & \omega_3=3 \\ & \left|G\left(\mathrm{j} \omega_3\right)\right|=\frac{1}{\sqrt{1+0.09 \times 9}}=0.74 \\ & \angle G\left(\mathrm{j} \omega_3\right)=-\tan ^{-1} 0.9=-42.0^{\circ} \end{aligned} \end{aligned}.
\begin{aligned} &n=5\\ &\begin{aligned} & \omega_5=5 \\ & \left|G\left(\mathrm{j} \omega_5\right)\right|=\frac{1}{\sqrt{1+0.09 \times 25}}=0.55 \\ & \angle G\left(\mathrm{j} \omega_5\right)=-\tan ^{-1} 1.5=-56.3 \end{aligned} \end{aligned}It is clear that high-frequency Fourier components are attenuated and phase shifted more than low-frequency Fourier components. The effect isto produce a rounding of the rising and falling edges of the square wave input signal. This is illustrated in Figure 23.29(b). The output signal has been obtained by adding together the attenuated and phase-shifted output Fourier components. This is possible because the system is linear.
