Aufgaben:Exercise 3.6: Transient Behavior - Revision history

Guenter at 16:22, 9 December 2021

2021-12-09T16:22:50Z

Guenter at 15:57, 9 December 2021

2021-12-09T15:57:22Z

Guenter at 15:49, 9 December 2021

2021-12-09T15:49:58Z

Guenter at 15:38, 9 December 2021

2021-12-09T15:38:02Z

Oezer: Oezer moved page Aufgaben:Exercise 3.6: Transient Behaviour to Aufgaben:Exercise 3.6: Transient Behavior

2021-10-25T08:03:38Z

Oezer moved page Exercise 3.6: Transient Behaviour to Exercise 3.6: Transient Behavior

Oezer at 14:24, 24 October 2021

2021-10-24T14:24:59Z

Oezer at 14:21, 24 October 2021

2021-10-24T14:21:53Z

Oezer at 14:20, 24 October 2021

2021-10-24T14:20:21Z

Oezer at 14:18, 24 October 2021

2021-10-24T14:18:56Z

Oezer at 14:16, 24 October 2021

2021-10-24T14:16:15Z

@@ Line 7: / Line 7: @@
 :$$c(t) = \cos(2\pi \cdot {t}/{T})
   \hspace{0.05cm} .$$
-In contrast,&nbsp; the causal cosine signal &nbsp;$c_{\rm K}(t)$&nbsp; (red curve)&nbsp; starts only at the turn-on instant&nbsp; $t = 0$:
+The causal&nbsp; $($German:&nbsp; "kausal" &nbsp; &rArr; &nbsp; subscript&nbsp; "K"$)$&nbsp; cosine signal &nbsp;$c_{\rm K}(t)$&nbsp; (red curve)&nbsp; starts only at the turn-on instant&nbsp; $t = 0$:
 :$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
    \end{array} \right.

@@ Line 133: / Line 133: @@
   {2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) =
   {0.02}/{T}\hspace{0.05cm}.$$
 [[File:P_ID1785__LZI_A_3_6_e.png|right|frame|Transient behavior of a causal cosine and a causal sine signal]]
@@ Line 143: / Line 142: @@
-The graph shows the output signals&nbsp; $y_{\rm C}(t)$&nbsp; and&nbsp; $y_{\rm CK}(t)$&nbsp; for cosine-shaped input on the left and the signals&nbsp; $y_{\rm S}(t)$&nbsp; and&nbsp; $y_{\rm SK}(t)$&nbsp; for sinusoidal input on the right. Note the transit time of&nbsp; $T/5$&nbsp; $($corresponding to the phase&nbsp; $72^\circ)$&nbsp; in both cases.
+The graph shows the output signals&nbsp; $y_{\rm C}(t)$&nbsp; and&nbsp; $y_{\rm CK}(t)$&nbsp; for cosine-shaped input on the left and the signals&nbsp; $y_{\rm S}(t)$&nbsp; and&nbsp; $y_{\rm SK}(t)$&nbsp; for sinusoidal input on the right.&nbsp; '''Note the transit time'''&nbsp; of&nbsp; $T/5$&nbsp; $($corresponding to the phase&nbsp; $72^\circ)$&nbsp; '''in both cases'''.
-*For&nbsp; $y_{\rm CK}(t)$&nbsp; the first wave peaks are smaller than $1$ to achieve the correct phase position of&nbsp; $y_{\rm C}(t)$&nbsp;.
+*For&nbsp; $y_{\rm CK}(t)$&nbsp; the first wave peaks are smaller than&nbsp; $1$&nbsp; to achieve the correct phase position of&nbsp; $y_{\rm C}(t)$&nbsp;.
-*In contrast, for&nbsp; $y_{\rm SK}(t)$&nbsp; the first wave peaks are greater than $1$ to achieve the correct phase position of&nbsp; $y_{\rm S}(t)$&nbsp;.
+*In contrast,&nbsp; for&nbsp; $y_{\rm SK}(t)$&nbsp; the first wave peaks are greater than&nbsp; $1$&nbsp; to achieve the correct phase position of&nbsp; $y_{\rm S}(t)$&nbsp;.
@@ Line 174: / Line 172: @@
   \hspace{0.05cm} .$$
-* Similarly, the following is obtained for the second part:
+* Similarly,&nbsp; the following is obtained for the second part:
 :$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
   \hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=
@@ Line 187: / Line 185: @@
   \hspace{0.05cm} .$$
-* Combining both parts and considering the numerical values of&nbsp; $p_{\rm x1}$&nbsp; and&nbsp; $p_{\rm x3}$,
+* Combining both parts and considering the numerical values of&nbsp; $p_{\rm x1}$&nbsp; and&nbsp; $p_{\rm x3}$:
 :$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}
   \cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}
@@ Line 201: / Line 199: @@
   \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
   \hspace{0.03cm}t/T}
-  \hspace{0.05cm} $$ is obtained.
+  \hspace{0.05cm} $$
 * Using Euler's theorem this can also be expressed as follows:
 :$$y_{1\hspace{-0.03cm}+2} (t) =  \frac { \cos(2\pi  {t}/{T}) + \pi \cdot \sin(2\pi  {t}/{T})} { {1 + \pi^2
   }}= y_{\rm C}(t)\hspace{0.05cm}.$$
-It can be seen that&nbsp; $y_{1\hspace{0.03cm}+2}(t)$&nbsp; is equal to the signal&nbsp; $y_{\rm C}(t)$&nbsp; computed in subtask&nbsp; '''(2)'''&nbsp;.
+:It can be seen that&nbsp; $y_{1\hspace{0.03cm}+2}(t)$&nbsp; is equal to the signal&nbsp; $y_{\rm C}(t)$&nbsp; computed in subtask&nbsp; '''(2)'''&nbsp;.
-* Finally, the following is obtained for the last residual:
+* Finally,&nbsp; the following is obtained for the last residual:
 :$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm
 x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm
@@ Line 222: / Line 220: @@
   \hspace{0.05cm} .$$
-* Thus, the output signal is as follows for a causal cosine signal applied to the input:
+* Thus,&nbsp; the output signal is as follows for a causal cosine signal applied to the input:
 :$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi  {t}/{T}) + \pi \cdot \sin(2\pi  {t}/{T})-{\rm
 e}^{\hspace{0.05cm}-2
@@ Line 232: / Line 230: @@
 :For comparison: &nbsp; The signal&nbsp; $y_{\rm C}(t)$&nbsp; has the value&nbsp; $0.303$ &nbsp; at this time.
-* In contrast to this, the following arises as a result in general and specifically at the time of the first maximum at&nbsp; $t = 0.45 \cdot T$ for the causal sinusoidal signal applied to the input:
+* In contrast to this,&nbsp; the following arises as a result in general and specifically at the time of the first maximum at&nbsp; $t = 0.45 \cdot T$&nbsp; for the causal sinusoidal signal applied to the input:
 :$$y_{\rm SK}(t) =  \frac { -\pi \cdot \cos(2\pi  {t}/{T}) + \sin(2\pi  {t}/{T})+\pi \cdot {\rm
 e}^{\hspace{0.05cm}-2

@@ Line 43: / Line 43: @@
-''Please note:''
+Please note:
 *The exercise belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
 *The computations for subtask&nbsp; '''(6)'''&nbsp; are bulky.
@@ Line 92: / Line 92: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Replacing the parameter&nbsp; $T$&nbsp; by&nbsp; $1/f_0$&nbsp; and&nbsp; $p$&nbsp; by&nbsp; ${\rm j} \cdot 2 \pi f$ in&nbsp; $H_{\rm L}(p)$&nbsp; the following is thus obtained for the frequency response in general or for&nbsp; $f_0 = 1 \ \rm MHz$:
+'''(1)'''&nbsp; Replacing the parameter&nbsp; $T$&nbsp; by&nbsp; $1/f_0$&nbsp; and &nbsp; $p$ &nbsp; by&nbsp; ${\rm j} \cdot 2 \pi f$ in&nbsp; $H_{\rm L}(p)$&nbsp; the following is obtained for the frequency response in general or for&nbsp; $f_0 = 1 \ \rm MHz$:
 :$$H(f) =  \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=
   \frac {f_0} { {\rm j} \cdot \pi f + f_0}
@@ Line 108: / Line 108: @@
 '''(2)'''&nbsp; The output signal&nbsp; $y_{\rm C}(t)$&nbsp; is attenuated by the factor&nbsp; $0.303$&nbsp; and delayed by&nbsp; $\tau \approx 72/360 \cdot T = T/5$&nbsp; compared to the input signal&nbsp; $c(t)$&nbsp;.
-Thus, this signal can also be described as follows:
+*Thus, this signal can also be described as follows:
 :$$y_{\rm C}(t) =  \frac { \cos(2\pi  {t}/{T}) + \pi \cdot \sin(2\pi  {t}/{T})} { {1 + \pi^2
   }}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow  \hspace{0.3cm} y_{\rm C}(t=0) =  \frac {1} { {1 + \pi^2
   }} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
-This signal is shown dotted in blue in the left graph for solution&nbsp; '''(5)'''&nbsp;.
+*This signal is shown dotted in blue in the left graph for solution&nbsp; '''(5)'''.
@@ Line 118: / Line 118: @@
 '''(3)'''&nbsp; The signal&nbsp; $y_{\rm S}(t)$&nbsp; is also smaller by the attenuation factor&nbsp; $0.303$&nbsp; and delayed by the time period&nbsp; $\tau = T/5$&nbsp; compared to&nbsp; $s(t)$&nbsp;.
-It can be described as follows:
+*It can be described as follows:
 :$$y_{\rm S}(t) =  \frac { -\pi \cdot \cos(2\pi  {t}/{T}) +  \sin(2\pi  {t}/{T})} { {1 + \pi^2
   }}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow  \hspace{0.3cm} y_{\rm S}(t=0) =  -\frac {\pi} { {1 + \pi^2
   }} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
-This signal is sketched dotted in blue in the right graph for solution&nbsp; '''(5)'''&nbsp;.
+*This signal is sketched dotted in blue in the right graph for solution&nbsp; '''(5)'''&nbsp;.
-'''(4)'''&nbsp; At&nbsp; $T_h$&nbsp; the impulse response&nbsp; $h(t)$&nbsp; should have decayed to&nbsp; $1\%$&nbsp; of the maximum value. Thus, the following holds:
+'''(4)'''&nbsp; At&nbsp; $T_h$&nbsp; the impulse response&nbsp; $h(t)$&nbsp; should have decayed to&nbsp; $1\%$&nbsp; of the maximum value.&nbsp;  Thus,&nbsp; the following holds:
 :$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2
   \hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}
@@ Line 135: / Line 135: @@
+[[File:P_ID1785__LZI_A_3_6_e.png|right|frame|Transient behavior of a causal cosine and a causal sine signal]]
-'''(5)'''&nbsp; The <u>statements 1 and 2</u> are correct:
+<br>'''(5)'''&nbsp; The <u>statements 1 and 2</u> are correct:
 *The causal signals&nbsp; $y_{\rm CK}(t)$&nbsp; and&nbsp; $y_{\rm SK}(t)$&nbsp; must be identical to zero for&nbsp; $t < 0$&nbsp;.
-*However, since the impulse response&nbsp; $h(t)$&nbsp; of the considered filter (nearly) vanishes for&nbsp; $t > T_h$,&nbsp; it does not matter whether the unbounded cosine signal&nbsp;  $c(t)$&nbsp; or the causal signal&nbsp; $c_{\rm K}(t)$ is applied to the input after the transient effect is over.
+*However,&nbsp; since the impulse response&nbsp; $h(t)$&nbsp; of the considered filter (nearly) vanishes for&nbsp; $t > T_h$,&nbsp; it does not matter whether the unbounded cosine signal&nbsp;  $c(t)$&nbsp; or the causal signal&nbsp; $c_{\rm K}(t)$&nbsp; is applied to the input after the transient effect is over.
 *The same is true for the sinusoidal signals: &nbsp; For&nbsp; $t >T_{ h}$,&nbsp; &nbsp; $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.
 The graph shows the output signals&nbsp; $y_{\rm C}(t)$&nbsp; and&nbsp; $y_{\rm CK}(t)$&nbsp; for cosine-shaped input on the left and the signals&nbsp; $y_{\rm S}(t)$&nbsp; and&nbsp; $y_{\rm SK}(t)$&nbsp; for sinusoidal input on the right. Note the transit time of&nbsp; $T/5$&nbsp; $($corresponding to the phase&nbsp; $72^\circ)$&nbsp; in both cases.

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves,<br>each causal]]
+[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves,&nbsp; each causal]]
-In this exercise, we consider a cosine signal &nbsp;$c(t)$&nbsp; with amplitude&nbsp; $1$&nbsp; and period $T = 1 \ \rm &micro; s$, which is defined for all times&nbsp; $t$&nbsp; $($in the range&nbsp; $ \pm \infty)$&nbsp;:
+In this exercise,&nbsp; we consider a cosine signal &nbsp;$c(t)$&nbsp; with amplitude&nbsp; $1$&nbsp; and period&nbsp; $T = 1 \ \rm &micro; s$,&nbsp; which is defined for all times&nbsp; $t$&nbsp; $($in the range&nbsp; $ \pm \infty)$&nbsp;:
 :$$c(t) = \cos(2\pi \cdot {t}/{T})
   \hspace{0.05cm} .$$
-In contrast, the causal cosine signal &nbsp;$c_{\rm K}(t)$&nbsp; (red curve) starts only at the turn-on instant&nbsp; $t = 0$:
+In contrast,&nbsp; the causal cosine signal &nbsp;$c_{\rm K}(t)$&nbsp; (red curve)&nbsp; starts only at the turn-on instant&nbsp; $t = 0$:
 :$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\
    \end{array} \right.
@@ Line 15: / Line 15: @@
 { t < 0\hspace{0.05cm}.}
 \end{array}$$
-Only the Fourier spectrum can be specified for the bilaterally unbounded signal &nbsp;$c(t)$&nbsp;:
+Only the Fourier spectrum can be specified for the bilaterally unbounded signal &nbsp;$c(t)$:
 :$$C(f) =  {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)
   \quad {\rm with} \quad f_0 =  {1}/{ T}= 1\,\,{\rm MHz.}$$
-On the contrary, for the causal cosine signal &nbsp;$c_{\rm K}(t)$&nbsp; the Laplace transform can also be specified:
+On the contrary,&nbsp; for the causal cosine signal &nbsp;$c_{\rm K}(t)$&nbsp; the Laplace transform can also be specified:
 :$$C_{\rm L}(p) =
   \frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
-Accordingly, the following holds for the Laplace transform of the causal sine function &nbsp;$s_{\rm K}(t)$:
+Accordingly,&nbsp; the following holds for the Laplace transform of the causal sine function &nbsp;$s_{\rm K}(t)$:
 :$$S_{\rm L}(p) =
   \frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$
@@ Line 27: / Line 27: @@
-The signals &nbsp;$c(t)$,  &nbsp;$c_{\rm K}(t)$, &nbsp;$s(t)$&nbsp; and &nbsp;$s_{\rm K}(t)$&nbsp; are applied to the input of a low-pass filter of first-order with the following transfer function (or impulse response):
+The signals &nbsp;$c(t)$,  &nbsp;$c_{\rm K}(t)$, &nbsp;$s(t)$&nbsp; and &nbsp;$s_{\rm K}(t)$&nbsp; are applied to the input of a low-pass filter of first-order with the following transfer function&nbsp; (or impulse response):
 :$$H_{\rm L}(p) =
   \frac {2 /T} { p + 2 /T} \quad
@@ Line 44: / Line 44: @@
 ''Please note:''
 *The exercise belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
 *The computations for subtask&nbsp; '''(6)'''&nbsp; are bulky.
 *For computing the signals &nbsp;$y_{\rm CK}(t)$&nbsp; and &nbsp;$y_{\rm SK}(t)$,&nbsp; for example the&nbsp; [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]&nbsp; can be used.
@@ Line 53: / Line 52: @@
 <quiz display=simple>
-{Compute the frequency response &nbsp;$H(f)$&nbsp; by magnitude and phase using &nbsp;$H_{\rm L}(p)$&nbsp;. &nbsp; What values are obtained for frequency &nbsp;$ f = f_0 = 1/T  = 1 \ \rm MHz$?
+{Compute the frequency response &nbsp;$H(f)$&nbsp; by magnitude and phase using &nbsp;$H_{\rm L}(p)$. &nbsp; What values are obtained for frequency &nbsp;$ f = f_0 = 1/T  = 1 \ \rm MHz$?
 |type="{}"}
 $|H(f = f_0)| \ = \ $  { 0.303 3% }
@@ Line 66: / Line 65: @@
-{Compute the output signal &nbsp;$y_{\rm S}(t)$ if the sine signal &nbsp;$s(t)$&nbsp; is applied to the filter input.&nbsp; What value is obtained for&nbsp; $t = 0$?
+{Compute the output signal &nbsp;$y_{\rm S}(t)$&nbsp; if the sine signal &nbsp;$s(t)$&nbsp; is applied to the filter input.&nbsp; What value is obtained for&nbsp; $t = 0$?
 |type="{}"}
 $y_{\rm S}(t = 0) \ = \ $  { -0.295--0.283 }
-{Determine the length of influence &nbsp;$T_h$&nbsp; of the filter impulse response, that is the time at which &nbsp;$h(t)$&nbsp; has decayed to &nbsp;$1\%$&nbsp; of the maximum value. Normalization to&nbsp; $T$.
+{Determine the influence  length &nbsp;$T_h$&nbsp; of the filter impulse response,&nbsp; that is the time at which &nbsp;$h(t)$&nbsp; has decayed to &nbsp;$1\%$&nbsp; of the maximum value.&nbsp; Normalization to&nbsp; $T$.
 |type="{}"}
 $T_h/T \ =  \ $  { 2.3 3% }

@@ Line 232: / Line 232: @@
 e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2
   }} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$
-:For comparison: &nbsp; The signal&nbsp; $y_{\rm C}(t)$&nbsp; has the value&nbsp; $0.303$ at this time.
+:For comparison: &nbsp; The signal&nbsp; $y_{\rm C}(t)$&nbsp; has the value&nbsp; $0.303$ &nbsp; at this time.
 * In contrast to this, the following arises as a result in general and specifically at the time of the first maximum at&nbsp; $t = 0.45 \cdot T$ for the causal sinusoidal signal applied to the input:

← Older revision	Revision as of 08:03, 25 October 2021
(No difference)

@@ Line 156: / Line 156: @@
   p_{\rm x3}= -{2}/{T}
   \hspace{0.05cm}$&nbsp;
-the following can be written for the Laplace transform of the signals&nbsp; $y_{\rm CK}(t)$&nbsp;:
+the following can be written for the Laplace transform of the signal&nbsp; $y_{\rm CK}(t)$&nbsp;:
 :$$Y_{\rm L}(p) =
   \frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}

@@ Line 147: / Line 147: @@
 The graph shows the output signals&nbsp; $y_{\rm C}(t)$&nbsp; and&nbsp; $y_{\rm CK}(t)$&nbsp; for cosine-shaped input on the left and the signals&nbsp; $y_{\rm S}(t)$&nbsp; and&nbsp; $y_{\rm SK}(t)$&nbsp; for sinusoidal input on the right. Note the transit time of&nbsp; $T/5$&nbsp; $($corresponding to the phase&nbsp; $72^\circ)$&nbsp; in both cases.
 *For&nbsp; $y_{\rm CK}(t)$&nbsp; the first wave peaks are smaller than&nbsp; $1$ &nbsp; to achieve the correct phase position of&nbsp; $y_{\rm C}(t)$&nbsp;.
-*In contrast, for&nbsp; $y_{\rm SK}(t)$&nbsp; the first wave peaks are greater than&nbsp; $1$ to achieve the correct phase position of&nbsp; $y_{\rm S}(t)$&nbsp;.
+*In contrast, for&nbsp; $y_{\rm SK}(t)$&nbsp; the first wave peaks are greater than&nbsp; $1$ &nbsp; to achieve the correct phase position of&nbsp; $y_{\rm S}(t)$&nbsp;.