A signal is assumed to be bandlimited to kHz. It is desired to filter this signal with an ideal bandpass filter that will pass the frequencies between kHz and kHz by a system for processing analog signals composed of a digital filter with frequency response sandwiched between an ideal A/D and an ideal D/A, both operating at sampling interval . 1. Determine the Nyquist sampling frequency, (in kHz), for the input signal. 2. Find the largest sampling period (in s) for which the overall system comprising A/D, digital filter and D/A realize the desired band pass filter.

Answers

Answer 1

Answer:

Hello your question is poorly written attached below is the complete question

answer :

1) 60 kHz

2)  Tmax  = ( 1 / 34000 ) secs

Explanation:

1) Determine the Nyquist sampling frequency, (in kHz), for the input signal.

F(s) = 2 * Fmax

Fmax = 30 kHz  ( since Xa(t) is band limited to 30 kHz )

∴ Nyquist sampling frequency ( F(s) ) = 2 * 30 = 60 kHz

2) Determine the largest sampling period (in s) .

Nyquist sampling period = 1 / Fs  = ( 1 / 60000 ) s

but there is some aliasing of the input signal ( minimum aliasing frequency > cutoff frequency of filter ) hence we will use the relationship below

=  2π - 2π * T * 30kHz  ≥  2π * T * 4kHz

∴ T ≤ [tex]\frac{1}{34kHz}[/tex]

largest sampling period ( Tmax ) = ( 1 / 34000 ) secs

A Signal Is Assumed To Be Bandlimited To KHz. It Is Desired To Filter This Signal With An Ideal Bandpass

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preguntas a parte o no???????

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guy above me is wrong

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Answers

Answer:

Explanation:

[tex]\text{The curve of the plot}[/tex] [tex]\mathbf{E_A,E_R, \ and \ E_N}[/tex] [tex]\text{can be seen in the attached diagram below}[/tex]

[tex]\text{From the plot}[/tex], [tex]\mathbf{r_o = 0.2 4nm \ and \ E_o =-5.3 eV}[/tex]

[tex]\mathbf{We \ knew \ that: E_N = E_A + E_R}[/tex]

[tex]\mathtt{GIven \ E_A = \dfrac{-1.436}{r}\ \ \ , E_R = \dfrac{7.32 \times 10^{-6}}{r^n} \ \ and \ \ n=8 }[/tex]

[tex]\mathtt{Then; E_N = -\dfrac{-1.436}{r}+ \dfrac{7.32\times 10^{-6}}{r^8}}[/tex]

[tex]\mathtt{Also; r_o = \Big( \dfrac{A}{nB} \Big)^{\dfrac{1}{1-n}}} \\ \\ \mathtt{ r_o = \Big( \dfrac{1.986}{8 \times 7.32\times 10^{-6}} \Big)^{\dfrac{1}{1-8}}} \\ \\ \mathbf{r_o = 0.236 nm}[/tex]

[tex]E_o = \dfrac{-1.436}{\Big[\dfrac{1.436}{8(732\times 10^{-6})}\Big]^{\dfrac{1}{1-8}}} + \dfrac{7.32 \times 10^{-6}}{\Big[ \dfrac{1.436}{8\times7.32 \times 10^{-6} } \Big]^{\dfrac{8}{1-8}}}[/tex]

[tex]\mathbf{E_o = -5.32 \ eV}[/tex]

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