The Fourier Transform and LTI Systems

Oct 13, 2025 by magicalceylontours.com 22 views

Introduction

In the field of signal processing, Linear Time-Invariant (LTI) systems play a crucial role in understanding how signals are transformed and modified. When dealing with LTI systems, it's essential to understand how they respond to different types of inputs, including periodic signals like the cosine function. In this article, we'll delve into the output of an LTI system for a cosine input and explore the implications of the given equation.

Understanding LTI Systems

Before we dive into the specifics of the cosine input, let's briefly review the basics of LTI systems. An LTI system is characterized by its impulse response, which is the output of the system when the input is the unit impulse function. The impulse response is denoted by the function $h[n]$ and is a fundamental property of LTI systems.

The Fourier Transform and LTI Systems

The Fourier Transform is a powerful tool for analyzing LTI systems. It allows us to represent the input and output signals in the frequency domain, making it easier to understand how the system responds to different frequencies. The Fourier Transform of the impulse response $h[n]$ is denoted by $H(e^{j\omega})$ , where $\omega$ is the angular frequency.

The Output of an LTI System for a Cosine Input

Now, let's focus on the output of an LTI system for a cosine input. The given equation states that the output $y[n]$ is given by:

cos[\omega_0n]\xrightarrow{h[n]} y[n]=|H(e^{j\omega_0})|\times cos[w_0n+\arg({H(e^{j\omega_0})})]

This equation suggests that the output of the LTI system is a scaled and phase-shifted version of the input cosine function. The scaling factor is given by the magnitude of the Fourier Transform of the impulse response at the frequency $\omega_0$ , denoted by $|H(e^{j\omega_0})|$ . The phase shift is determined by the argument of the Fourier Transform of the impulse response at the frequency $\omega_0$ , denoted by $\arg({H(e^{j\omega_0})})$ .

Interpretation of the Equation

To better understand the implications of this equation, let's break it down into its components. The term $|H(e^{j\omega_0})|$ represents the magnitude of the system's response at the frequency $\omega_0$ . This value indicates how much the system amplifies or attenuates the input signal at that frequency. The term $\arg({H(e^{j\omega_0})})$ represents the phase shift introduced by the system at the frequency $\omega_0$ . This value indicates how much the system delays or advances the input signal at that frequency.

Doubt in the Equation

However, there seems to be a discrepancy in the equation. If we substitute $x[n]=cos[n]$ into the equation, we get:

cos[n]\xrightarrow{h[n]} y[n]=|H(e^{j1})|\times cos[n+\arg({H(e^{j1})})]

But for $H(e^{j1})$ , the magnitude is 1 and the argument is 0, which means that the output should be simply $cos[n]$ . This seems to contradict the given equation, which suggests that the output is a scaled and phase-shifted version of the input cosine function.

Conclusion

In conclusion, the output of an LTI system for a cosine input is a complex function that depends on the system's impulse response and the frequency of the input signal. The given equation provides a useful framework for understanding how LTI systems respond to periodic signals like the cosine function. However, there seems to be a discrepancy in the equation, which requires further investigation to resolve.

Future Work

To resolve the discrepancy in the equation, further analysis is needed. One possible approach is to examine the properties of the Fourier Transform and its relationship to the impulse response of the LTI system. Additionally, numerical simulations can be used to verify the accuracy of the equation and identify any potential errors.

References

[1] Oppenheim, A. V., & Willsky, A. S. (1997). Signals and Systems. Prentice Hall.
[2] Proakis, J. G., & Manolakis, D. G. (1996). Digital Signal Processing: Principles, Algorithms, and Applications. Prentice Hall.
[3] Strang, G. (1999). Linear Algebra and Its Applications. Thomson Learning.

Appendix

For the sake of completeness, let's provide a brief derivation of the equation for the output of an LTI system for a cosine input.

Let $x[n]=cos[n]$ be the input signal, and let $h[n]$ be the impulse response of the LTI system. The output signal $y[n]$ is given by the convolution of the input signal and the impulse response:

y[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]

Substituting $x[n]=cos[n]$ into this equation, we get:

y[n]=\sum_{k=-\infty}^{\infty}cos[k]h[n-k]

Using the trigonometric identity $cos[k]=\frac{e^{jk}+e^{-jk}}{2}$ , we can rewrite the equation as:

y[n]=\frac{1}{2}\sum_{k=-\infty}^{\infty}(e^{jk}+e^{-jk})h[n-k]

Now, let's consider the Fourier Transform of the impulse response $h[n]$ . The Fourier Transform of $h[n]$ is denoted by $H(e^{j\omega})$ , where $\omega$ is the angular frequency. Using the linearity property of the Fourier Transform, we can write:

H(e^{j\omega})=\sum_{k=-\infty}^{\infty}h[k]e^{-jk\omega}

Substituting this expression into the equation for $y[n]$ , we get:

y[n]=\frac{1}{2}\sum_{k=-\infty}^{\infty}(e^{jk}+e^{-jk})\sum_{m=-\infty}^{\infty}h[m]e^{-jm\omega}

Using the convolution property of the Fourier Transform, we can rewrite this equation as:

y[n]=\frac{1}{2}\sum_{m=-\infty}^{\infty}h[m]\sum_{k=-\infty}^{\infty}(e^{jk}+e^{-jk})e^{-jm\omega}

Now, let's consider the term $\sum_{k=-\infty}^{\infty}(e^{jk}+e^{-jk})e^{-jm\omega}$ . Using the trigonometric identity $e^{jk}+e^{-jk}=2cos(k)$ , we can rewrite this term as:

\sum_{k=-\infty}^{\infty}(e^{jk}+e^{-jk})e^{-jm\omega}=2\sum_{k=-\infty}^{\infty}cos(k)e^{-jm\omega}

Using the orthogonality property of the cosine function, we can rewrite this term as:

2\sum_{k=-\infty}^{\infty}cos(k)e^{-jm\omega}=2\pi\delta(\omega-m)

Substituting this expression into the equation for $y[n]$ , we get:

y[n]=\pi\sum_{m=-\infty}^{\infty}h[m]\delta(\omega-m)

Now, let's consider the term $\sum_{m=-\infty}^{\infty}h[m]\delta(\omega-m)$ . Using the sifting property of the delta function, we can rewrite this term as:

\sum_{m=-\infty}^{\infty}h[m]\delta(\omega-m)=H(e^{j\omega})

Substituting this expression into the equation for $y[n]$ , we get:

y[n]=\pi H(e^{j\omega})

Now, let's consider the term $H(e^{j\omega})$ . Using the definition of the Fourier Transform, we can rewrite this term as:

H(e^{j\omega})=\sum_{k=-\infty}^{\infty}h[k]e^{-jk\omega}

Using the linearity property of the Fourier Transform, we can rewrite this term as:

H(e^{j\omega})=|H(e^{j\omega})|e^{j\arg(H(e^{j\omega}))}

Substituting this expression into the equation for $y[n]$ , we get:

y[n]=\pi |H(e^{j\omega})|e^{j\arg(H(e^{j\omega}))}

Now, let's consider the term $\pi |H(e^{j\omega})|e^{j\arg(H(e^{j\omega}))}$ . Using the definition of the Fourier Transform, we can rewrite this term as:

\pi |H(e^{j\omega})|e^{j\arg(H(e^{j\omega}))}=\pi |H(e^{j\omega_0})|e^{j\arg(H(e^{j\omega_0<br/> **Q&A: Doubt in Output of an LTI System for a Cosine Input** ===========================================================

Q: What is the output of an LTI system for a cosine input?

A: The output of an LTI system for a cosine input is a complex function that depends on the system's impulse response and the frequency of the input signal. The given equation provides a useful framework for understanding how LTI systems respond to periodic signals like the cosine function.

Q: What is the equation for the output of an LTI system for a cosine input?

A: The equation for the output of an LTI system for a cosine input is given by: