Aufgaben:Exercise 5.3Z: Non-Recursive Filter

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Nonrecursive filter

Consider the adjacent non-recursive filter with the filter coefficients

$$a_0 = 1,\quad a_1 = 2,\quad a_2 = 1.$$

We are looking for the respective output sequences  $\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle$ when the following value sequences are applied to the input:

• the "equal sequence"

$$\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{g_\nu } \right\rangle = \left\langle {\;1,\;1,\;1,\;1,\;1,\;1,\;1,\;1,\;\text{...} } \hspace{0.05cm} \right\rangle .$$

• the "sinusoidal sequence" with period  $T_0 = 4 \cdot T_{\rm A}$:

$$\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{s_\nu } \right\rangle = \left\langle {\;0,\;1,\;0, - 1,\;0,\;1,\;0, - 1,\;\text{...} } \hspace{0.05cm}\right\rangle .$$

Notes:

• The exercise belongs to the chapter  Digital Filters in this book.
• Reference is also made to some chapters in the book  Signal Representation.

Questions

1

What is the filter impulse response  $h(t)$?  At what time  $\nu \cdot T_{\rm A}$  does the impulse response have its maximum?

$\nu \ = \ $

2

Calculate the frequency response  $H(f)$.  What is the value at $f = 0$?

$H(f = 0) \ = \ $

3

What is the output sequence  $\left\langle \hspace{0.05cm} {y_\nu } \hspace{0.05cm} \right\rangle$  for the DC sequence  $\left\langle \hspace{0.05cm} {g_\nu } \hspace{0.05cm}\right\rangle$  at its input?  Interpret this result considering the last subtask.
What is the output value for  $\nu = 4 $?

$y_4 \ = \ $

4

What output sequence  $\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle$  results for the sine sequence $\left\langle \hspace{0.05cm}{s_\nu } \hspace{0.05cm}\right\rangle$ at its input?  What output value results for  $\nu = 4 $?

$y_4 \ = \ $

Solution

Solution

(1)   The impulse response is:   $h(t) = \delta (t) + 2 \cdot \delta ( {t - T_{\rm A} } ) + \delta ( {t - 2T_{\rm A} } ).$

• The maximum is at  $T_{\rm A}$,   ⇒   $\underline{\nu = 1}$.

(2)   The frequency response  $H(f)$  is the Fourier transform of the impulse response  $h(t)$.

• The impulse response shifted to the left by  $T_{\rm A}$ 

$$h\hspace{0.05cm}'(t) = \delta ( {t + T_{\rm A} } ) + 2 \cdot \delta ( t ) + \delta ( {t - T_{\rm A} } )$$

is symmetric at  $t= 0$  and accordingly has the purely real frequency response

$$H\hspace{0.05cm}'(f) = 2\big [ {1 + \cos ( {2{\rm{\pi }}fT_{\rm A} } )} \big ].$$

• By applying the shifting theorem, it further follows:

$$H(f) = 2\big [ {1 + \cos ( {2{\rm{\pi }}fT_{\rm A} } )} \big ] \cdot {\rm{e}}^{ - {\rm{j}}2{\rm{\pi }}fT_{\rm A} } .$$

• Consequently, the value of the frequency response at frequency $f=0$ is $H(f = 0)\hspace{0.15cm}\underline{ = 4}$.

(3)   The discrete-time convolution of the input sequence  $\left\langle \hspace{0.05cm}{g_\nu } \hspace{0.05cm} \right\rangle$  with the impulse response  $\left\langle \hspace{0.05cm}{h_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{1, 2, 1 } \hspace{0.05cm}\right\rangle$  results in

$$\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;3,\;4,\;4,\;4,\;4,\;4,\;4,\;4,\;4,\;4,\; \text{...} \;} \right\rangle $$

• In particular,  $y_4\hspace{0.15cm}\underline{ = 4}$.
• With the exception of the values  $y_0$  and  $y_1$  (transient), we also obtain a synchronous sequence at the output with the constant value 4:

$$y(t) = H( {f = 0} ) \cdot x( t ) = 4 \cdot 1 = 4.$$

(4)   Analogous to subtask  (3),  we now obtain by shifting, weighting with  $a_1$,  $a_2$,  $a_3$  and subsequent superposition:

$$\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{\;0,\;1,\;2,\;0,\; - 2,\;0,\;2,\;0,\; - 2,\;0,\;...\;}\hspace{0.05cm} \right\rangle .$$

• Thus, the value we are looking for is  $y_4\hspace{0.15cm}\underline{ = -2}$.

Another solution:

• The input sequence  $\left\langle \hspace{0.05cm}{s_\nu }\hspace{0.05cm} \right\rangle$  is sinusoidal with period  $4 \cdot T_{\rm A}$.  Accordingly, the fundamental frequency is  $f_0 = 1/(4 \cdot T_{\rm A})$.
• At this frequency, the frequency response  $H(f)$  has the following value according to subtask  (2): 

$$H( {f = f_0 } ) = 2\big[ {1 + \cos ( {{{\rm{\pi }}}/{2}} )} \big] \cdot {\rm{e}}^{ - {\rm{j\pi /2}}} = 2 \cdot {\rm{e}}^{ - {\rm{j\pi /2}}} .$$

• Leaving the transient  $($completed at  $t = T_{\rm A})$  out of consideration, the following relationship between the input and output signals is obtained with  $\tau = T_{\rm A}$   $($phase:   $90^\circ)$: 

$$y(t) = 2 \cdot x( {t - T_{\rm A} } ).$$

• That means:   The sine function becomes the function "minus-cosine" with the amplitude 2.