Eigenfunctions of LTI Systems: Complex Exponentials
3.2 Eigenfunctions of LTI Systems: Complex Exponentials
In Signals and Systems, we seek “basic signals” with two crucial properties:
Complex exponentials (\(e^{st}\) and \(z^n\)) fit these criteria perfectly.
A signal for which the system output is a (possibly complex) constant times the input is called an eigenfunction of the system.
For a system \(\mathcal{H}\), if: \[ \mathcal{H}\{x(t)\} = y(t) = \lambda x(t) \]
then \(x(t)\) is an eigenfunction of the system, and \(\lambda\) is the corresponding eigenvalue.
Continuous-Time
Input: \(x(t) = e^{st}\)
Output: \(y(t) = H(s) e^{st}\)
Here, \(e^{st}\) is the eigenfunction, and \(H(s)\) is the corresponding eigenvalue.
Discrete-Time
Input: \(x[n] = z^n\)
Output: \(y[n] = H(z) z^n\)
Here, \(z^n\) is the eigenfunction, and \(H(z)\) is the corresponding eigenvalue.
Let an LTI system have impulse response \(h(t)\).
Input: \(x(t) = e^{st}\).
The output \(y(t)\) is given by the convolution integral: \[ y(t) = \int_{-\infty}^{+\infty} h(\tau) x(t-\tau) d \tau \]
Substitute \(x(t-\tau) = e^{s(t-\tau)}\): \[ y(t) = \int_{-\infty}^{+\infty} h(\tau) e^{s(t-\tau)} d \tau \] \[ y(t) = \int_{-\infty}^{+\infty} h(\tau) e^{st} e^{-s\tau} d \tau \]
Since \(e^{st}\) is independent of \(\tau\), we can pull it out of the integral: \[ y(t) = e^{st} \int_{-\infty}^{+\infty} h(\tau) e^{-s\tau} d \tau \]
Thus, the output is: \[ y(t) = H(s) e^{st} \] where the eigenvalue \(H(s)\) is: \[ H(s) = \int_{-\infty}^{+\infty} h(\tau) e^{-s\tau} d \tau \]
Similarly, for a discrete-time LTI system with impulse response \(h[n]\).
Input: \(x[n] = z^n\).
The output \(y[n]\) is given by the convolution sum: \[ y[n] = \sum_{k=-\infty}^{+\infty} h[k] x[n-k] \]
Substitute \(x[n-k] = z^{n-k}\): \[ y[n] = \sum_{k=-\infty}^{+\infty} h[k] z^{n-k} \] \[ y[n] = \sum_{k=-\infty}^{+\infty} h[k] z^n z^{-k} \]
Since \(z^n\) is independent of \(k\), we can pull it out of the sum: \[ y[n] = z^n \sum_{k=-\infty}^{+\infty} h[k] z^{-k} \]
Thus, the output is: \[ y[n] = H(z) z^n \] where the eigenvalue \(H(z)\) is: \[ H(z) = \sum_{k=-\infty}^{+\infty} h[k] z^{-k} \]
The eigenfunction property, combined with the linearity of LTI systems (superposition), significantly simplifies analysis.
Consider an input composed of a sum of complex exponentials: \[ x(t) = a_1 e^{s_1 t} + a_2 e^{s_2 t} + a_3 e^{s_3 t} \]
By superposition, the response to the sum is the sum of individual responses:
Summing these up, the total output \(y(t)\) is: \[ y(t) = a_1 H(s_1) e^{s_1 t} + a_2 H(s_2) e^{s_2 t} + a_3 H(s_3) e^{s_3 t} \]
If the input to a continuous-time LTI system is a linear combination of complex exponentials:
Input Signal \[ x(t) = \sum_{k} a_k e^{s_k t} \]
Output Signal \[ y(t) = \sum_{k} a_k H(s_k) e^{s_k t} \]
The output is simply a linear combination of the same complex exponentials, but each scaled by its corresponding eigenvalue \(H(s_k)\).
Similarly, for a discrete-time LTI system:
Input Signal \[ x[n] = \sum_{k} a_k z_k^n \]
Output Signal \[ y[n] = \sum_{k} a_k H(z_k) z_k^n \]
The output is also a linear combination of the same complex exponential sequences, each scaled by its corresponding eigenvalue \(H(z_k)\).
While \(s\) and \(z\) can be arbitrary complex numbers, Fourier analysis focuses on specific values:
Continuous-Time We restrict \(s\) to purely imaginary values: \[ s = j\omega \] This focuses on complex exponentials of the form \(e^{j\omega t}\).
Discrete-Time We restrict \(z\) to values of unit magnitude: \[ z = e^{j\omega} \] This focuses on complex exponentials of the form \(e^{j\omega n}\).
Consider an LTI system where the output \(y(t)\) is a time-shifted version of the input \(x(t)\): \[ y(t)=x(t-3). \] This system represents a pure time delay of 3 units.
Part 1: Response to a single complex exponential input
Input: \(x(t) = e^{j2t}\).
To find the output, we directly substitute \(x(t)\) into the system equation: \[ y(t) = e^{j2(t-3)} = e^{-j6} e^{j2t}. \]
Comparing \(y(t) = e^{-j6} e^{j2t}\) with the eigenfunction form \(y(t) = H(s) e^{st}\), we can identify: The eigenvalue \(H(j2) = e^{-j6}\).
Let’s verify the eigenvalue \(H(s)\) by first determining the system’s impulse response \(h(t)\).
For a system defined by \(y(t)=x(t-3)\), the impulse response is a delayed impulse function: \(h(t) = \delta(t-3)\).
Now, we use the formula for \(H(s)\) derived earlier: \[ H(s) = \int_{-\infty}^{+\infty} h(\tau) e^{-s\tau} d \tau \] Substitute \(h(\tau) = \delta(\tau-3)\): \[ H(s) = \int_{-\infty}^{+\infty} \delta(\tau-3) e^{-s\tau} d \tau \]
Due to the sifting property of the Dirac delta function, this integral evaluates to: \[ H(s) = e^{-s(3)} = e^{-3s} \]
Finally, we evaluate \(H(s)\) at \(s=j2\) (the frequency of our input \(x(t)=e^{j2t}\)): \[ H(j2) = e^{-j3(2)} = e^{-j6} \] This result precisely matches the eigenvalue we found by direct substitution on the previous slide!
Consider a more complex input signal composed of a sum of two cosine waves: \[ x(t) = \cos(4t) + \cos(7t) \]
Direct approach: From \(y(t)=x(t-3)\), the output is simply: \[ y(t) = \cos(4(t-3)) + \cos(7(t-3)) \]
Using Eigenfunctions & Superposition: First, we express \(x(t)\) using Euler’s formula to decompose it into complex exponentials: \[ x(t) = \frac{1}{2}e^{j4t} + \frac{1}{2}e^{-j4t} + \frac{1}{2}e^{j7t} + \frac{1}{2}e^{-j7t} \]
We already found the system’s eigenvalue \(H(j\omega)\) for a time delay of 3: \(H(j\omega) = e^{-j3\omega}\).
Now, we apply the eigenfunction property to each complex exponential component of \(x(t)\):
For \(\frac{1}{2}e^{j4t}\): The eigenvalue is \(H(j4) = e^{-j3(4)} = e^{-j12}\).
Output component: \(\frac{1}{2} H(j4) e^{j4t} = \frac{1}{2} e^{-j12} e^{j4t} = \frac{1}{2} e^{j(4t-12)} = \frac{1}{2} e^{j4(t-3)}\)
For \(\frac{1}{2}e^{-j4t}\): The eigenvalue is \(H(-j4) = e^{-j3(-4)} = e^{j12}\).
Output component: \(\frac{1}{2} H(-j4) e^{-j4t} = \frac{1}{2} e^{j12} e^{-j4t} = \frac{1}{2} e^{-j(4t-12)} = \frac{1}{2} e^{-j4(t-3)}\)
For \(\frac{1}{2}e^{j7t}\): The eigenvalue is \(H(j7) = e^{-j3(7)} = e^{-j21}\).
Output component: \(\frac{1}{2} H(j7) e^{j7t} = \frac{1}{2} e^{-j21} e^{j7t} = \frac{1}{2} e^{j(7t-21)} = \frac{1}{2} e^{j7(t-3)}\)
For \(\frac{1}{2}e^{-j7t}\): The eigenvalue is \(H(-j7) = e^{-j3(-7)} = e^{j21}\).
Output component: \(\frac{1}{2} H(-j7) e^{-j7t} = \frac{1}{2} e^{j21} e^{-j7t} = \frac{1}{2} e^{-j(7t-21)} = \frac{1}{2} e^{-j7(t-3)}\)
Summing these individual outputs, we get the total output \(y(t)\): \[ y(t) = \frac{1}{2} e^{j4(t-3)} + \frac{1}{2} e^{-j4(t-3)} + \frac{1}{2} e^{j7(t-3)} + \frac{1}{2} e^{-j7(t-3)} \] Applying Euler’s formula in reverse (recall \(\cos(\theta) = \frac{e^{j\theta} + e^{-j\theta}}{2}\)): \[ y(t) = \cos(4(t-3)) + \cos(7(t-3)) \] This result precisely matches the direct calculation, demonstrating the power of the eigenfunction approach!
Key Takeaways:
What’s Next?
This lays the essential foundation for Fourier Analysis: