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<a href=":Category:Problem solving">Practice Problem</a> on Discrete-time Fourier transform computation
Compute the discrete-time Fourier transform of the following signal:
<img _fckfakelement="true" _fck_mw_math=" x[n]= \sin \left( \frac{2 \pi }{100} n \right) " src="/rhea/images/math/b/f/0/bf0f97ec20b83c8416e3cd5d95395388.png" />
(Write enough intermediate steps to fully justify your answer.)
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Answer 1
<img _fckfakelement="true" _fck_mw_math="x[n]=\sin \left( \frac{2 \pi}{100} \right)" src="/rhea/images/math/2/2/3/2231dd05d43da978a3499a964d730f15.png" />
<img _fckfakelement="true" _fck_mw_math="x[n] = \frac{1}{2j} \left( e^{ \frac{j2 \pi}{100n}}-e^{- \frac{j2 \pi}{100n}} \right)" src="/rhea/images/math/8/3/6/83609c9889b65fcd96711aed18f6111c.png" />
<img _fckfakelement="true" _fck_mw_math="X_(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n}" src="/rhea/images/math/2/b/d/2bd6e9f4f8c6f4d0c8ba6b2119288a9f.png" />
<img _fckfakelement="true" _fck_mw_math="X_(\omega) = \frac{1}{2j} \left( \sum_{n=-\infty}^{+\infty} e^{ \frac{j2 \pi} {100} n} e^{-j\omega n} - \sum_{n=-\infty}^{+\infty} e^{\frac{-j2 \pi} {100} n} e^{-j\omega n} \right)" src="/rhea/images/math/7/f/9/7f9ab2bb6f04aadec1f5789b52ed7e47.png" />
<img _fckfakelement="true" _fck_mw_math="X_(\omega) = \frac{\pi}{j} \left( \delta \left({\omega - \frac{2 \pi}{100}}\right) - \delta \left({\omega + \frac{2 \pi}{100}}\right) \right) by DTFT table" src="/rhea/images/math/6/1/8/61814381fa1ee34cad6f25374d24430a.png" />
Answer 2
First, write the original function as: <img _fckfakelement="true" _fck_mw_math="x[n] = \frac{1}{2j} \left( e^{ \frac{j2 \pi}{100n}}-e^{- \frac{j2 \pi}{100n}} \right)" src="/rhea/images/math/8/3/6/83609c9889b65fcd96711aed18f6111c.png" />
Then, for w = [-pi, pi]
<img _fckfakelement="true" _fck_mw_math="X_(\omega) = \frac{1}{2j} \left( \sum_{n=-\infty}^{+\infty} e^{ \frac{j2 \pi} {100} n} e^{-j\omega n} - \sum_{n=-\infty}^{+\infty} e^{\frac{-j2 \pi} {100} n} e^{-j\omega n} \right)" src="/rhea/images/math/7/f/9/7f9ab2bb6f04aadec1f5789b52ed7e47.png" />
<img _fckfakelement="true" _fck_mw_math="X_(\omega) = \frac{100}{2j} \left( \delta \left( \frac{100}{2pi}\omega - 1 \right) + \left( \frac{100}{2pi}\omega + 1 \right) \right)" src="/rhea/images/math/f/0/c/f0c3875a9621aabf1286b15e89b68e9f.png" />
which is really is:
<img _fckfakelement="true" _fck_mw_math="X_(\omega) = rep_2pi \frac{50}{j} \left( \delta \left( \frac{100}{2pi}\omega - 1 \right) + \left( \frac{100}{2pi}\omega + 1 \right) \right)" src="/rhea/images/math/3/7/b/37b3b44a22cf3b61840a7c8bb4486eec.png" />
Answer 3
We can separate the equation to the following function
<img _fckfakelement="true" _fck_mw_math="x[n]=\frac{1}{2 j} \left( e^\frac{j 2 \pi n}{100} - e^\frac{- j 2 \pi n}{100} \right) " src="/rhea/images/math/0/5/b/05ba7655493b0c4e3b694ba6fac0539c.png" />
Because based on Fourier transform equation,
<img _fckfakelement="true" _fck_mw_math="X_(\omega) = \sum_{n = -\infty}^{\infty} x[n] e^{-j \omega n}" src="/rhea/images/math/a/7/9/a79fed22e488f9ab5773eadadc46bcb0.png" />
Substitute in x[n]
<img _fckfakelement="true" _fck_mw_math="X_(\omega) = \frac{1}{2 j} \left( \sum_{n = -\infty}^{\infty} e^{ \frac{j2 \pi n} {100} } e^{-j\omega n} - \sum_{n = -\infty}^{ \infty} e^{\frac{-j2 \pi n} {100} } e^{-j\omega n} \right)" src="/rhea/images/math/b/2/8/b283311397523aab2fbcfa322f3b759f.png" />
From Discrete Fourier Transform pair,
<img _fckfakelement="true" _fck_mw_math=" x[n] = e^{-j\omega_0 n} " src="/rhea/images/math/1/1/9/119b4dfb4f6bcf5deb0663acaa69ca03.png" /> DTFT to <img _fckfakelement="true" _fck_mw_math=" X_(\omega) = 2 \pi \sum_{n = -\infty}^{ \infty} \delta \left( \omega-\omega_0 - 2\pi l \right) " src="/rhea/images/math/b/2/5/b25d82705497c4e62e1d068138dae962.png" />
Hence, the function will be
<img _fckfakelement="true" _fck_mw_math=" X_(\omega) = \frac{\pi}j \left( \sum_{n = -\infty}^{ \infty} \delta \left( \omega-\omega_0 - 2\pi l \right) - \sum_{n = -\infty}^{ \infty} \delta \left( \omega+\omega_0 - 2\pi l \right) \right) " src="/rhea/images/math/5/2/c/52cb905ce8f853f34e9fb0f152c902f0.png" />
<img _fckfakelement="true" _fck_mw_math="x[n]=\sin \left( \frac{2\pi}{100} n \right)" src="/rhea/images/math/b/f/0/bf0f97ec20b83c8416e3cd5d95395388.png" />
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