A signal $2 \cos \left(\frac{2\pi}{3}t\right) - \cos (\pi t)$ is the input to an LTI system with the transfer function $H(s) = e^s + e^{-s}$. If $C_k$ denotes the $k^{th}$ coefficient in the exponential Fourier series of the output signal, then $C_3$ is equal to
To determine the coefficient \( C_3 \) in the exponential Fourier series of the output signal, we first analyze the input to the LTI system and the given transfer function.
The input signal is:
\(x(t) = 2 \cos \left(\frac{2\pi}{3} t\right) - \cos(\pi t)\)
The cosine terms can be rewritten using Euler's formula:
\(2 \cos \left(\frac{2\pi}{3} t\right) = e^{j \frac{2\pi}{3} t} + e^{-j \frac{2\pi}{3} t}\)
\(\cos(\pi t) = \frac{1}{2}(e^{j \pi t} + e^{-j \pi t})\)
Substituting these, the input signal in terms of exponentials is:
\(x(t) = \left(e^{j \frac{2\pi}{3} t} + e^{-j \frac{2\pi}{3} t}\right) - \frac{1}{2}(e^{j \pi t} + e^{-j \pi t})\)
The input signal is comprised of frequency components at \(\frac{2\pi}{3}\) and \(\pi\).
The transfer function of the system is:
\(H(s) = e^s + e^{-s}\)
The key to solving this is recognizing that the transfer function will affect each frequency component of the input signal. The exponential Fourier series coefficients of the output signal will be obtained by multiplying the Fourier series coefficients of the input signal by the transfer function evaluated at the respective frequencies.
Evaluating the transfer function at these frequencies:
- For \(\omega = \frac{2\pi}{3}\), \(H(j\frac{2\pi}{3}) = e^{j \frac{2\pi}{3}} + e^{-j \frac{2\pi}{3}}\)
- For \(\omega = \pi\), \(H(j\pi) = e^{j\pi} + e^{-j\pi}\\)
Compute the values:
\(H(j\frac{2\pi}{3}) = e^{j \frac{2\pi}{3}} + e^{-j \frac{2\pi}{3}} = 2 \cos \left(\frac{2\pi}{3}\right) = -1\)
\(H(j\pi) = e^{j\pi} + e^{-j\pi} = 2 \cos (\pi) = -2\)
After passing the signal through the system, the output signal's frequency components become:
\(Y(f) = -1 \cdot \left(e^{j \frac{2\pi}{3} t} + e^{-j \frac{2\pi}{3} t}\right) -2 \cdot \frac{1}{2} (e^{j \pi t} + e^{-j \pi t})\)
Therefore, the Fourier series coefficients of the output signal are:
- For \(k = \pm 1\), \( C_{\pm 1} = -1 \)
- For \(k = \pm 3\), \( C_{\pm 3} = -1 \)
This gives us:
\(C_3 = -1 \cdot 1 = 1\)
Hence, the Fourier coefficient \(C_3\) of the output signal is 1.
Important Questions from Fourier Series
The Fourier series representation of a square wave is shown in the figure below. The fluctuations seen near $x = \pm 1$ are named after which one of the following scientists?
The Fourier series expansion of the function $f(x) = |\cos(x)|$ in the interval $(-\pi, \pi)$ is $\alpha + \beta \left[ \frac{1}{3}\cos(2x) - \frac{1}{15}\cos(4x) + \dots \right]$.
The values of $\alpha$ and $\beta$ respectively are ______.
Periodic function $f(x)$ is given below. $$f(x) = \begin{cases} -1, & \text{when } -\pi < x < 0 \\ 1, & \text{when } 0 < x < \pi \end{cases} ; f(x + 2\pi) = f(x)$$ The CORRECT option representing the Fourier series expansion of $f(x)$ is:
A time-limited waveform $g(x)$ is specified as follows: $$g(x) = \begin{cases} -k, & -\pi < x \le 0 \\ +k, & 0 < x \le \pi \\ 0, & \text{otherwise} \end{cases}$$ A new waveform $f(x)$ is constructed from $g(x)$ as follows: $$f(x) = \sum_{m=-\infty}^{\infty} g(x + 2\pi m), \quad \text{for all } x \in \mathbb{R}$$ The sum of the coefficients of the third harmonics of the sine and cosine terms in the trigonometric Fourier series expansion of $f(x)$ is $\frac{2}{3\pi}$.
What is the value of $k$?
The effective value of the current for the Sawtooth waveform as shown is:
