The output of a continuous-time, linear time-invariant system is denoted by $T\left\{ {x\left( t \right)} \right\}$ where $x\left( t \right)$ is the input signal. A signal $z\left( t \right)$ is called eigen-signal of the system , when $T\left\{ {z\left( t \right)} \right\} = \gamma z\left( t \right)$, where $\gamma $ is a complex number, in general, and is called an eigenvalue of $T$. Suppose the impulse response of the system $T$ is real and even. Which of the following statements is TRUE?

Question

Prepp Team · Accepted Answer

Understanding Eigen-signals in LTI Systems

The question asks us to identify the behavior of cosine and sine signals when passed through a linear time-invariant (LTI) system whose impulse response $h(t)$ is both real and even. An eigen-signal $z(t)$ of a system $T$ is defined by the property that the system's output is just a scaled version of the input signal itself:

$$ T\left\{ {z\left( t \right)} \right\} = \gamma z\left( t \right) $$

Here, $\gamma$ is a scalar constant known as the eigenvalue associated with the eigen-signal $z(t)$. For LTI systems, the output $y(t)$ when the input is $x(t)$ is given by the convolution of the input signal with the system's impulse response $h(t)$:

$$ y(t) = T\left\{ {x\left( t \right)} \right\} = x\left( t \right) * h\left( t \right) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau $$

Analyzing Cosine and Sine Signals

We need to check if $cos(t)$ and $sin(t)$ satisfy the eigen-signal condition for an LTI system with a real and even impulse response $h(t)$. A key property is that if $h(t)$ is real and even, its Fourier Transform, $H(\omega)$, is also real and even.

Case 1: Input Signal is $cos(t)$

Let the input signal be $x(t) = cos(t)$. The output $y(t)$ is:

$$ y(t) = cos(t) * h(t) $$

Using the Fourier Transform properties, the Fourier Transform of the output $Y(\omega)$ is the product of the Fourier Transforms of the input $X(\omega)$ and the impulse response $H(\omega)$:

$$ Y(\omega) = X(\omega) H(\omega) $$

The Fourier Transform of $cos(t)$ is $X(\omega) = \pi[\delta(\omega - 1) + \delta(\omega + 1)]$. Since $h(t)$ is real and even, $H(\omega)$ is real and even, meaning $H(\omega) = H(-\omega)$.

Therefore,

$$ Y(\omega) = \pi[\delta(\omega - 1) + \delta(\omega + 1)] H(\omega) $$ $$ Y(\omega) = \pi H(1) \delta(\omega - 1) + \pi H(-1) \delta(\omega + 1) $$

Since $H(\omega)$ is even, $H(1) = H(-1)$. Let $H(1) = \gamma_1$.

$$ Y(\omega) = \pi \gamma_1 [\delta(\omega - 1) + \delta(\omega + 1)] $$

This $Y(\omega)$ is proportional to the Fourier Transform of $cos(t)$. This implies that the output signal $y(t)$ is a scaled version of $cos(t)$:

$$ y(t) = \gamma_1 cos(t) $$

Thus, $cos(t)$ is an eigen-signal with eigenvalue $\gamma_1 = H(1)$.

Case 2: Input Signal is $sin(t)$

Let the input signal be $x(t) = sin(t)$. The output $y(t)$ is:

$$ y(t) = sin(t) * h(t) $$

The Fourier Transform of $sin(t)$ is $X(\omega) = \pi j[\delta(\omega + 1) - \delta(\omega - 1)]$. Again, $H(\omega)$ is real and even.

$$ Y(\omega) = X(\omega) H(\omega) = \pi j[\delta(\omega + 1) - \delta(\omega - 1)] H(\omega) $$ $$ Y(\omega) = \pi j H(1) [\delta(\omega + 1) - \delta(\omega - 1)] - \pi j H(-1) [\delta(\omega - 1) - \delta(\omega + 1)] $$

Since $H(1) = H(-1)$, let $H(1) = \gamma_2$.

$$ Y(\omega) = \pi j \gamma_1 [\delta(\omega + 1) - \delta(\omega - 1)] + \pi j \gamma_1 [\delta(\omega + 1) - \delta(\omega - 1)] $$ $$ Y(\omega) = \pi j \gamma_1 [\delta(\omega + 1) - \delta(\omega - 1)] $$

This $Y(\omega)$ is proportional to the Fourier Transform of $sin(t)$. This implies that the output signal $y(t)$ is a scaled version of $sin(t)$:

$$ y(t) = \gamma_1 sin(t) $$

Thus, $sin(t)$ is also an eigen-signal with eigenvalue $\gamma_2 = H(1)$.

Comparing Eigenvalues

From the analysis above, the eigenvalue for $cos(t)$ is $\gamma_1 = H(1)$ and the eigenvalue for $sin(t)$ is $\gamma_2 = H(1)$.

Therefore, both $cos(t)$ and $sin(t)$ are eigen-signals of the LTI system, and they share the same (identical) eigenvalue, which is the value of the system's frequency response $H(\omega)$ evaluated at $\omega = 1$ (or $\omega = -1$).

Conclusion

Based on the derivation, both $cos(t)$ and $sin(t)$ are eigen-signals for the described system, and they possess identical eigenvalues. This matches option 4.