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Question

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

The correct answer is
1

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.

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Important Questions from Fourier Series

  1. If we use the Fourier transform ϕ(x, y) =  \(\int {{{\rm{e}}^{{\rm{ikx}}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}}\)  to solve the partial differential equation  \({\rm{ - }}\frac{{{\partial ^2}ϕ \left( {x,y} \right)}}{{\partial {y^2}}}\, - \,\frac{1}{{{y^2}}}\frac{{{\partial ^2}ϕ \left( {x,y} \right)}}{{\partial {x^2}}} + \frac{{{m^2}}}{{{y^2}}}ϕ \left( {x,y} \right) = 0\)  in the half-plane {(x, y) : -∞ < x < ∞, 0 < y < ∞} the Fourier modes ϕ k(y) depend on y as y α  and y β . The values of α and β are  

  2. When a time-domain signal is converted into its Fourier representation, which of the following is/are conserved?

    I. Energy

    II. Power

  3. The trigonometric Fourier series of a periodic time function can have

  4. The Fourier series expansion of x3 in the interval −1 ≤ x < 1 with periodic continuation has

  5. The Fourier series to represent x-x2 for –π ≤ x ≤ π is given by \(x - {x^2} = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}cosnx + \mathop \sum \limits_{n = 1}^\infty {b_n}sinnx\)

    The value of a0 (round off to two decimal places), is
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