- #1

jashua

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The energy and the power contents of a signal x(t), denoted by [itex]E_x[/itex] and [itex]P_x[/itex], respectively, are defined as

[itex](1) E_x = \int ^{\infty}_{-\infty} |x(t)|^2 dt[/itex]

[itex](2) P_x = lim_{T\rightarrow \infty} \frac{1}{T} \int ^{T/2}_{-T/2} |x(t)|^2 dt[/itex]

Let us use the discrete time (sampled) signal, with sampling period [itex]T_s[/itex]. Then, the first problem is to show that these definitions are equivalent to the following ones:

[itex](3) E_x = T_s \sum ^{\infty}_{n=-\infty} |x[n]|^2 [/itex]

[itex](4) P_x = lim_{N\rightarrow \infty} \frac{1}{2N+1} \sum ^{N}_{n=-N} |x[n]|^2 [/itex]

Now, if we use the FFT, that is, if the length of the sequence is finite and the sequence is repeated. Then, the next problem is to show that these definitions (#1 and #2) are equivalent to the following ones (i think only one period should be considered for [itex]E_x[/itex]):

[itex](5) E_x = T_s \sum ^{N-1}_{n=0} |x[n]|^2 [/itex]

[itex](6) P_x = \frac{1}{N} \sum ^{N-1}_{n=0} |x[n]|^2 [/itex]

My work to solve these questions are as follows:

Let us define the sampled waveform [itex]x_s(t)[/itex]:

[itex]x_s(t) = x(t) \sum ^{\infty}_{n=-\infty}\delta (t-nTs).[/itex]

[itex] = \sum ^{\infty}_{n=-\infty}x(nT_s)\delta (t-nTs).[/itex]

[itex] = \sum ^{\infty}_{n=-\infty}x[n]\delta (t-nTs).[/itex]

Then, the Fourier transform of [itex]x_s(t)[/itex], [itex]X_s(f)[/itex] is

[itex]X_s(f) = \sum ^{\infty}_{n=-\infty}x[n]e^{-j2\pi f n T_s}.[/itex]

which is nothing but the DFT of the discrete time sequence x[n].

On the other hand, we can express [itex]X_s(f)[/itex] in terms of [itex]X(f)[/itex] (using Poissson's sum formula) as follows:

[itex]X_s(f) = X(f) \ast \frac{1}{T_s} \sum ^{\infty}_{n=-\infty}\delta (n-\frac{n}{T_s}).[/itex]

[itex] = \frac{1}{T_s} \sum ^{\infty}_{n=-\infty}X(f-\frac{n}{T_s}).[/itex]

where [itex]\ast[/itex] is the convolution operator.

Hence, we have

[itex]X_s(f)=\frac{1}{T_s}X(f)[/itex], [itex]|f|<W[/itex].

Then, the energy spectral density is found to be [itex]\epsilon_x (f) = |X(f)|^2 = {T_s}^2 |X_s(f)|^2 [/itex]. Now we can find [itex]E_x[/itex] using the following formula:

[itex]E_x = \int ^{\infty}_{-\infty} \epsilon_x (f) df[/itex]

[itex] = {T_s}^2 \int ^{W}_{-W} |X_s(f)|^2 df[/itex]

[itex] = {T_s}^2 \int ^{W}_{-W} \sum ^{\infty}_{n=-\infty}x[n]e^{-j2\pi f n T_s} \sum ^{\infty}_{m=-\infty}x[m]^\ast e^{j2\pi f m T_s}df[/itex]

[itex] = {T_s}^2 \sum ^{\infty}_{n=-\infty}x[n] \sum ^{\infty}_{m=-\infty}x[m]^\ast \int ^{W}_{-W} e^{j2\pi f (m-n) T_s}df[/itex]

[itex] = {T_s}^2 \sum ^{\infty}_{n=-\infty}x[n] \sum ^{\infty}_{m=-\infty}x[m]^\ast 2W sinc(2WT_s(m-n)) [/itex]

[itex] = {T_s}^2 2W \sum ^{\infty}_{n=-\infty} |x[n]|^2 [/itex]

where [itex]x[m]^\ast[/itex] is the conjugate of [itex]x[m][/itex]. If [itex]T_s=\frac{1}{2W}[/itex], then we have shown the definitions #1 and #3 are equivalent.In order to show that the definitions #2 and #4 are equivalent, let [itex]T=(2N+1)T_s[/itex] such that

[itex]x_{s_T}(t)= \sum ^{N}_{n=-N}x[n]\delta (t-nTs)[/itex]

Then, we have,

[itex] lim_{T\rightarrow \infty}X_{s_T}(f) = lim_{N\rightarrow \infty} \frac{1}{T_s} \sum ^{N}_{n=-N}X(f-\frac{n}{T_s}) [/itex]

And hence,

[itex] lim_{T\rightarrow \infty} X_{s_T}(f)=\frac{1}{T_s}X(f)[/itex], [itex]|f|<W[/itex].

Substituting [itex]X(f)=T_s X_{s_T}(f)[/itex] into the definition # 1 (after applying Parseval relation), we get

[itex]P_x = lim_{T\rightarrow \infty} \frac{1}{T} \int ^{T/2}_{-T/2} |x(t)|^2 dt[/itex]

[itex]= lim_{T\rightarrow \infty} \frac{1}{T} \int ^{\infty}_{-\infty}|x(t)|^2 dt[/itex]

[itex]= lim_{T\rightarrow \infty} \frac{1}{T} \int ^{\infty}_{-\infty}|X(f)|^2 df[/itex]

[itex]= lim_{T\rightarrow \infty} \frac{1}{T} {T_s}^2 \int ^{W}_{-W}|X_{s_T}(f)|^2 df[/itex]

Now substituting [itex]X_{s_T}(f)=\sum ^{N}_{n=-N}x[n]e^{-j2\pi f n T_s}[/itex], and [itex]T=(2N+1)T_s[/itex], we get

[itex]P_x = lim_{N\rightarrow \infty} \frac{1}{(2N+1)T_s} {T_s}^2 \int ^{W}_{-W} \sum ^{N}_{n=-N}x[n]e^{-j2\pi f n T_s} \sum ^{N}_{m=-N}x[m]^\ast e^{j2\pi f m T_s} df[/itex]

[itex] = lim_{N\rightarrow \infty} \frac{1}{(2N+1)T_s} {T_s}^2 2W \sum ^{N}_{n=-N} |x[n]|^2[/itex]

If [itex]T_s=\frac{1}{2W}[/itex], then,

[itex] P_x= lim_{N\rightarrow \infty} \frac{1}{2N+1} \sum ^{N}_{n=-N} |x[n]|^2[/itex]

Hence we have shown the definitions #2 and #4 are equivalent.

Am I correct up to this point?

I'm still trying to get some satisfactory solutions to show the relation between the definitions #1 and # 5 and the definitions #2 and # 6 and waiting for some help.

[itex](1) E_x = \int ^{\infty}_{-\infty} |x(t)|^2 dt[/itex]

[itex](2) P_x = lim_{T\rightarrow \infty} \frac{1}{T} \int ^{T/2}_{-T/2} |x(t)|^2 dt[/itex]

Let us use the discrete time (sampled) signal, with sampling period [itex]T_s[/itex]. Then, the first problem is to show that these definitions are equivalent to the following ones:

[itex](3) E_x = T_s \sum ^{\infty}_{n=-\infty} |x[n]|^2 [/itex]

[itex](4) P_x = lim_{N\rightarrow \infty} \frac{1}{2N+1} \sum ^{N}_{n=-N} |x[n]|^2 [/itex]

Now, if we use the FFT, that is, if the length of the sequence is finite and the sequence is repeated. Then, the next problem is to show that these definitions (#1 and #2) are equivalent to the following ones (i think only one period should be considered for [itex]E_x[/itex]):

[itex](5) E_x = T_s \sum ^{N-1}_{n=0} |x[n]|^2 [/itex]

[itex](6) P_x = \frac{1}{N} \sum ^{N-1}_{n=0} |x[n]|^2 [/itex]

My work to solve these questions are as follows:

Let us define the sampled waveform [itex]x_s(t)[/itex]:

[itex]x_s(t) = x(t) \sum ^{\infty}_{n=-\infty}\delta (t-nTs).[/itex]

[itex] = \sum ^{\infty}_{n=-\infty}x(nT_s)\delta (t-nTs).[/itex]

[itex] = \sum ^{\infty}_{n=-\infty}x[n]\delta (t-nTs).[/itex]

Then, the Fourier transform of [itex]x_s(t)[/itex], [itex]X_s(f)[/itex] is

[itex]X_s(f) = \sum ^{\infty}_{n=-\infty}x[n]e^{-j2\pi f n T_s}.[/itex]

which is nothing but the DFT of the discrete time sequence x[n].

On the other hand, we can express [itex]X_s(f)[/itex] in terms of [itex]X(f)[/itex] (using Poissson's sum formula) as follows:

[itex]X_s(f) = X(f) \ast \frac{1}{T_s} \sum ^{\infty}_{n=-\infty}\delta (n-\frac{n}{T_s}).[/itex]

[itex] = \frac{1}{T_s} \sum ^{\infty}_{n=-\infty}X(f-\frac{n}{T_s}).[/itex]

where [itex]\ast[/itex] is the convolution operator.

Hence, we have

[itex]X_s(f)=\frac{1}{T_s}X(f)[/itex], [itex]|f|<W[/itex].

Then, the energy spectral density is found to be [itex]\epsilon_x (f) = |X(f)|^2 = {T_s}^2 |X_s(f)|^2 [/itex]. Now we can find [itex]E_x[/itex] using the following formula:

[itex]E_x = \int ^{\infty}_{-\infty} \epsilon_x (f) df[/itex]

[itex] = {T_s}^2 \int ^{W}_{-W} |X_s(f)|^2 df[/itex]

[itex] = {T_s}^2 \int ^{W}_{-W} \sum ^{\infty}_{n=-\infty}x[n]e^{-j2\pi f n T_s} \sum ^{\infty}_{m=-\infty}x[m]^\ast e^{j2\pi f m T_s}df[/itex]

[itex] = {T_s}^2 \sum ^{\infty}_{n=-\infty}x[n] \sum ^{\infty}_{m=-\infty}x[m]^\ast \int ^{W}_{-W} e^{j2\pi f (m-n) T_s}df[/itex]

[itex] = {T_s}^2 \sum ^{\infty}_{n=-\infty}x[n] \sum ^{\infty}_{m=-\infty}x[m]^\ast 2W sinc(2WT_s(m-n)) [/itex]

[itex] = {T_s}^2 2W \sum ^{\infty}_{n=-\infty} |x[n]|^2 [/itex]

where [itex]x[m]^\ast[/itex] is the conjugate of [itex]x[m][/itex]. If [itex]T_s=\frac{1}{2W}[/itex], then we have shown the definitions #1 and #3 are equivalent.In order to show that the definitions #2 and #4 are equivalent, let [itex]T=(2N+1)T_s[/itex] such that

[itex]x_{s_T}(t)= \sum ^{N}_{n=-N}x[n]\delta (t-nTs)[/itex]

Then, we have,

[itex] lim_{T\rightarrow \infty}X_{s_T}(f) = lim_{N\rightarrow \infty} \frac{1}{T_s} \sum ^{N}_{n=-N}X(f-\frac{n}{T_s}) [/itex]

And hence,

[itex] lim_{T\rightarrow \infty} X_{s_T}(f)=\frac{1}{T_s}X(f)[/itex], [itex]|f|<W[/itex].

Substituting [itex]X(f)=T_s X_{s_T}(f)[/itex] into the definition # 1 (after applying Parseval relation), we get

[itex]P_x = lim_{T\rightarrow \infty} \frac{1}{T} \int ^{T/2}_{-T/2} |x(t)|^2 dt[/itex]

[itex]= lim_{T\rightarrow \infty} \frac{1}{T} \int ^{\infty}_{-\infty}|x(t)|^2 dt[/itex]

[itex]= lim_{T\rightarrow \infty} \frac{1}{T} \int ^{\infty}_{-\infty}|X(f)|^2 df[/itex]

[itex]= lim_{T\rightarrow \infty} \frac{1}{T} {T_s}^2 \int ^{W}_{-W}|X_{s_T}(f)|^2 df[/itex]

Now substituting [itex]X_{s_T}(f)=\sum ^{N}_{n=-N}x[n]e^{-j2\pi f n T_s}[/itex], and [itex]T=(2N+1)T_s[/itex], we get

[itex]P_x = lim_{N\rightarrow \infty} \frac{1}{(2N+1)T_s} {T_s}^2 \int ^{W}_{-W} \sum ^{N}_{n=-N}x[n]e^{-j2\pi f n T_s} \sum ^{N}_{m=-N}x[m]^\ast e^{j2\pi f m T_s} df[/itex]

[itex] = lim_{N\rightarrow \infty} \frac{1}{(2N+1)T_s} {T_s}^2 2W \sum ^{N}_{n=-N} |x[n]|^2[/itex]

If [itex]T_s=\frac{1}{2W}[/itex], then,

[itex] P_x= lim_{N\rightarrow \infty} \frac{1}{2N+1} \sum ^{N}_{n=-N} |x[n]|^2[/itex]

Hence we have shown the definitions #2 and #4 are equivalent.

Am I correct up to this point?

I'm still trying to get some satisfactory solutions to show the relation between the definitions #1 and # 5 and the definitions #2 and # 6 and waiting for some help.

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