Eigenfunctions Revisited 243
4.2 Eigenfunctions Revisited 243
the delay. Find the frequency response of the ideal communication system, and use it to determine the steady-state response when the delay caused by the system is τ = 3 sec, and the input is x(t) =
2 cos(4t − π/4).
Solution
The impulse response of the ideal system is h(t) = δ(t − τ ) where τ is the delay of the transmission. In fact, the output according to the convolution integral gives
y(t) =
δ(ρ −τ) x(t − ρ)dρ = x(t − τ ) | {z }
0 h(ρ)
as expected. Let us then find the frequency response of the ideal communication system. According to the eigenvalue property, if the input is x(t)
j =e t 0 , then the output is j y(t) t =e 0 H( j
but also
y(t)
= x(t − τ ) = e j 0 ( t −τ )
so that comparing these equations we have that
H( j 0 ) =e −jτ 0
For a generic frequency 0 ≤ < ∞, we would get
H( j) =e −jτ
which is a complex function of , with a unity magnitude |H( j)| = 1, and a linear phase ∠H( j) = −τ . This system is called an all-pass system, since it allows all frequency components of the input to go through with a phase change only.
Consider the case when τ = 3, and that we input into this system x(t) = 2 cos(4t − π/4), then H( j) = 1e −j3 , so that the output in the steady state is
y(t) = 2|H( j4)| cos(4t − π/4 + ∠H( j4)) = 2 cos(4(t − 3) − π/4) = x(t − 3)
where we used H( j4) = 1e −j12 (i.e., |H( j4)| = 1 and ∠H( j4) = 12). ■
■ Example 4.3
Although there are better methods to compute the frequency response of a system represented by
a differential equation, the eigenfunction property can be easily used for that. Consider the RC
C H A P T E R 4: Frequency Analysis: The Fourier Series
circuit shown in Figure 4.1 where the input is
v s ( t) = 1 + cos(10,000t)
with components of low frequency, = 0, and of large frequency, = 10,000 rad/sec. The output v c ( t) is the voltage across the capacitor in steady state. We wish to find the frequency response of this circuit to verify that it is a low-pass filter (it allows low-frequency components to go through, but filters out high-frequency components).
Solution
Using Kirchhoff’s voltage law, this circuit is represented by a first-order differential equation,
dv c ( t) v s ( t) =v c ( t) + dt
Now, if the input is v s ( t) =e jt , for a generic frequency , then the output is v c ( t) =e jt H( j). Replacing these in the differential equation, we have
so that
1 H( j) =
1 + j or the frequency response of the filter for any frequency . The magnitude of H( j) is
1 |H( j)| = √
which is close to one for small values of the frequency, and tends to zero when the frequency values are large—the characteristics of a low-pass filter.
For the input
v s ( t) = 1 + cos(10,000t) = cos(0t) + cos(10,000t)
(i.e., it has a zero frequency component and a 10,000-rad/sec frequency component) using Euler’s identity, we have that
v s ( t) = 1 + 0.5 e j10,000t +e −j10,000t
and the steady-state output of the circuit is
v c ( t)
= 1H( j0) + 0.5H( j10,000)e j10,000t + 0.5H(−j10,000)e −j10,000t
cos(10,000t − π/2) ≈ 1