Difference between revisions of "Aufgaben:Exercise 1.4: Low-Pass Filter of 2nd Order"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain}} |
− | [[File: | + | [[File:EN_LZI_A_1_4.png|right|frame|Low-pass filter of first order (top) <br>and of second order (bottom)]] |
− | In [[Aufgaben:1 | + | In [[Aufgaben:Exercise_1.1:_Simple_Filter_Functions|Exercise 1.1]] and [[Aufgaben:Exercise_1.1Z:_Low-Pass_Filter_of_1st_and_2nd_Order|Exercise 1.1Z]] of the chapter [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain|System Description in Frequency Domain]] the so-called "RC low-pass filters" were described in the frequency domain. Now, the time domain representation is elaborated on. |
− | + | The circuit with the input signal $x(t)$ and the output signal $y_1(t)$ is a low-pass filter of first order with frequency response | |
− | $$H_{\rm 1}(f) = \frac{1}{1+{\rm j}\cdot f/f_{\rm 0}}.$$ | + | :$$H_{\rm 1}(f) = \frac{1}{1+{\rm j}\cdot f/f_{\rm 0}}.$$ |
− | + | Here, $f_0 = 1/(2πRC)$ indicates the 3dB cut-off frequency. If a Dirac-shaped signal $x(t) = δ(t)$ is applied to the input, the signal $y_1(t)$ is given by the system output according to the sketch in the middle. | |
− | + | The relationship between the system parameters $R$ (resistance), $C$ (capacitance) and $T$ is (see [[Aufgaben:Exercise_1.3Z:_Exponentially_Decreasing_Impulse_Response|Exercise 1.3Z]]): | |
− | $$T = \frac{1}{\omega_{\rm 0}}= \frac{1}{2\pi f_{\rm 0}} = R C.$$ | + | :$$T = \frac{1}{\omega_{\rm 0}}= \frac{1}{2\pi f_{\rm 0}} = R \cdot C.$$ |
− | + | For numerical calculations $T = 1 \ \rm ms$ shall be used in the following. | |
− | + | The circuit below with input $x(t)$ and output $y_2(t)$ represents a low-pass filter of second order: | |
− | $$H_{\rm 2}(f) = [H_{\rm 1}(f)]^2 =\frac{1}{(1+{\rm j}\cdot f/f_{\rm 0})^2}.$$ | + | :$$H_{\rm 2}(f) = \big[H_{\rm 1}(f)\big]^2 =\frac{1}{(1+{\rm j}\cdot f/f_{\rm 0})^2}.$$ |
− | + | *The impulse response associated with $H_2(f)$ is $h_2(t)$. | |
+ | *The system parameter $f_0$ no longer indicates the 3dB cut-off frequency for a low-pass filter of second or higher order. | ||
+ | *Furthermore, it should be noted that the two RC elements must be decoupled in order to achieve impedance matching. | ||
+ | *An operational amplifier, for example, is suitable for this. However, this hint is not relevant for the solution of this task. | ||
− | |||
− | '' | + | |
− | $$\int u \cdot {\rm e}^{a \cdot \hspace{0.03cm} u} \hspace{0.1cm}{\rm d}u = | + | |
+ | |||
+ | |||
+ | |||
+ | |||
+ | ''Please note:'' | ||
+ | *The exercise belongs to the chapter [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain|System Description in Time Domain]]. | ||
+ | *The following indefinite integral is given: | ||
+ | :$$\int u \cdot {\rm e}^{a \cdot \hspace{0.03cm} u} \hspace{0.1cm}{\rm d}u = | ||
\frac{{\rm e}^{\hspace{0.03cm}a \cdot \hspace{0.03cm} u}}{a^2} \cdot (a \cdot u -1).$$ | \frac{{\rm e}^{\hspace{0.03cm}a \cdot \hspace{0.03cm} u}}{a^2} \cdot (a \cdot u -1).$$ | ||
+ | |||
+ | |||
− | === | + | |
+ | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {State the impulse response $h_1(t)$ . At what time $t_1$ has $h_1(t)$ decreased to half of its maximum value? |
|type="{}"} | |type="{}"} | ||
− | $t_1$ | + | $t_1\ = \ $ { 0.693 5% } $\ \rm ms$ |
− | { | + | {What is the output signal $y_1(t)$ for $x(t) = T · h_1(t)$? What are the signal values at times $t = 0$ and $t = T$ ? |
|type="{}"} | |type="{}"} | ||
− | $y_1(t = 0) =$ { 0 } | + | $y_1(t = 0) \ = \ $ { 0. } |
− | $y_1(t = T) =$ { 0.368 5% } | + | $y_1(t = T) \ = \ $ { 0.368 5% } |
− | { | + | {Compute the impulse response $h_2(t)$ considering the result of '''(2)'''. At what time $t_2$ is $h_2(t=t_2)$ maximum? |
|type="{}"} | |type="{}"} | ||
− | $t_2$ | + | $t_2 \ = \ $ { 1 } $\rm ms$ |
− | $h_2(t = t_2) =$ { 368 } 1/s | + | $h_2(t = t_2) \ = \ $ { 368 3% } $\rm 1/s$ |
− | { | + | {What is the output signal $y_2(t)$ if a step function $x(t) = {\rm 2 \hspace{0.05cm}V} · γ(t)$ is applied to the input? What are the signal values at $t = T$ and $t = 5T$ ? |
|type="{}"} | |type="{}"} | ||
− | $y_2(t = T) =$ { 0.528 5% } V | + | $y_2(t = T) \ = \ $ { 0.528 5% } $\rm V$ |
− | $y_2(t = 5T) =$ { 1.919 5% } V | + | $y_2(t = 5T) \ = \ $ { 1.919 5% } $\rm V$ |
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''1 | + | '''(1)''' By definition the impulse response is equal to the output signal if a Dirac delta function of weight $1$ is applied to the input. |
− | $$h_1(t) = y_1(t) ={1}/{T} \cdot {\rm e}^{-t/T}.$$ | + | *According to the [[Aufgaben:Exercise_1.3Z:_Exponentially_Decreasing_Impulse_Response|solution of Exercise 1.3Z]] and the sketch above the following holds: |
− | + | :$$h_1(t) = y_1(t) ={1}/{T} \cdot {\rm e}^{-t/T}.$$ | |
− | $$h_1(t_{\rm 1}) ={1}/{T} \cdot {\rm e}^{-t_{\rm 1}/T} = \frac{1}{2T} | + | *For the time $t_1$ the following shall hold: |
+ | :$$h_1(t_{\rm 1}) ={1}/{T} \cdot {\rm e}^{-t_{\rm 1}/T} = \frac{1}{2T} | ||
\hspace{0.5cm}\Rightarrow\hspace{0.5cm}{t_{\rm 1}}/{T} = {\rm ln}(2)\hspace{0.1cm} \Rightarrow \hspace{0.1cm}t_{\rm 1} = 0.693 \cdot T \hspace{0.15cm}\underline{= {\rm 0.693\,\,ms}}.$$ | \hspace{0.5cm}\Rightarrow\hspace{0.5cm}{t_{\rm 1}}/{T} = {\rm ln}(2)\hspace{0.1cm} \Rightarrow \hspace{0.1cm}t_{\rm 1} = 0.693 \cdot T \hspace{0.15cm}\underline{= {\rm 0.693\,\,ms}}.$$ | ||
− | |||
− | |||
− | |||
− | + | ||
− | $$\begin{align*}h_1(\tau) & = \frac{1}{T} \cdot {\rm e}^{-\tau/T}\hspace{0.1cm} \Rightarrow \hspace{0.1cm}h_1(-\tau) = \frac{1}{T} \cdot {\rm e}^{\tau/T}\hspace{0.1cm} \Rightarrow \hspace{0.1cm}h_1(t-\tau) = \frac{1}{T} \cdot {\rm e}^{(-t+\tau)/T} \\ & \Rightarrow\hspace{0.5cm} | + | |
+ | '''(2)''' The input signal $x(t)$ is an exponentially decreasing impulse like the impulse response $h_1(t)$ but dimensionless. | ||
+ | [[File:P_ID830__LZI_A_1_4b.png |right|frame| Illustration of the convolution]] | ||
+ | *Thus, according to the convolution theorem: | ||
+ | |||
+ | :$$y_1(t) = x (t) * h_1 (t) = T \cdot \big[ h_1 (t) * h_1 (t) \big].$$ | ||
+ | |||
+ | *The convolution is illustrated here for a specific time $t$ by a sketch. | ||
+ | |||
+ | *The following is obtained after renaming the variables: | ||
+ | :$$\begin{align*}h_1(\tau) & = \frac{1}{T} \cdot {\rm e}^{-\tau/T}\hspace{0.1cm} \Rightarrow \hspace{0.1cm}h_1(-\tau) = \frac{1}{T} \cdot {\rm e}^{\tau/T}\hspace{0.1cm} \Rightarrow \hspace{0.1cm}h_1(t-\tau) = \frac{1}{T} \cdot {\rm e}^{(-t+\tau)/T} \\ & \Rightarrow\hspace{0.5cm} | ||
y_1(t) = T \cdot \int\limits_{ - \infty }^{ + \infty } h_1 ( {t - \tau } )\cdot {h_1 ( \tau )} \hspace{0.1cm}{\rm d}\tau.\end{align*}$$ | y_1(t) = T \cdot \int\limits_{ - \infty }^{ + \infty } h_1 ( {t - \tau } )\cdot {h_1 ( \tau )} \hspace{0.1cm}{\rm d}\tau.\end{align*}$$ | ||
− | + | *For $τ < 0$ the impulse response is $h_1(τ) = 0$. For $τ > t$ the first convolution operand vanishes (see sketch). From this it follows that: | |
− | $$y_1(t) = T \cdot \frac{1}{T^2}\cdot \int_{ 0 }^{ t } {\rm | + | :$$y_1(t) = T \cdot \frac{1}{T^2}\cdot \int_{ 0 }^{ t } {\rm |
e}^{(-t+\tau)/T} \cdot {\rm e}^{-\tau/T} \hspace{0.1cm}{\rm d}\tau = \frac{1}{T}\cdot \int_{ 0 }^{ t } {\rm e}^{-t/T} \hspace{0.1cm}{\rm d}\tau .$$ | e}^{(-t+\tau)/T} \cdot {\rm e}^{-\tau/T} \hspace{0.1cm}{\rm d}\tau = \frac{1}{T}\cdot \int_{ 0 }^{ t } {\rm e}^{-t/T} \hspace{0.1cm}{\rm d}\tau .$$ | ||
− | + | *Now, it can be observed that the integrand is independent of the integration variable $τ$. Consequently, the following holds: | |
− | $$y_1(t) = | + | :$$y_1(t) = {t}/{T} \cdot {\rm e}^{-t/T}\hspace{0.1cm} \Rightarrow |
\hspace{0.1cm} y_1(t =0) \hspace{0.15cm}\underline{=0}; \hspace{0.5cm}y_1(t =T)={\rm e}^{-1} \hspace{0.15cm}\underline{=0.368}.$$ | \hspace{0.1cm} y_1(t =0) \hspace{0.15cm}\underline{=0}; \hspace{0.5cm}y_1(t =T)={\rm e}^{-1} \hspace{0.15cm}\underline{=0.368}.$$ | ||
− | '''3 | + | |
− | + | '''(3)''' Due to the interrelationship $H_2(f) = H_1(f) · H_1(f)$ the following holds for the impulse response: $h_2(t) = h_1 (t) * h_1 (t).$ | |
− | + | *Except for the additional constant factor $1/T$ the same result as in subtask '''(2)''' is obtained: | |
− | $$h_2(t) = | + | :$$h_2(t) ={t}/{T^2} \cdot {\rm e}^{-t/T}.$$ |
− | + | *The maximum value is obtained by setting the derivative to zero: | |
− | $$ | + | :$$\frac{{\rm d} h_2(t)}{{\rm d}t} = \frac{1}{T^2} \cdot {\rm e}^{-t/T} \cdot \left( 1 - {t}/{T}\right) = 0 \hspace{0.5cm} \Rightarrow \hspace{0.5cm} t_{\rm 2} \hspace{0.15cm}\underline{= T = 1\,\,{\rm ms}} \hspace{0.5cm}\Rightarrow \hspace{0.5cm} h_2(t_{\rm 2}) =\frac{{\rm e}^{-1}}{T} =\frac{0.368}{1\,\,{\rm ms}} \hspace{0.15cm}\underline{ = {\rm 368 \hspace{0.1cm} 1/s}}.$$ |
+ | |||
− | '''4 | + | '''(4)''' In general or with the result from '''(3)''' the following applies to the step response: |
− | $${\rm \sigma_2}(t) = \int_{ 0 }^{ t } {h_2 ( \tau )} \hspace{0.1cm}{\rm d}\tau = | + | :$${\rm \sigma_2}(t) = \int_{ 0 }^{ t } {h_2 ( \tau )} \hspace{0.1cm}{\rm d}\tau = |
\frac{1}{T^2} \cdot \int_{ 0 }^{ t } \tau \cdot {\rm e}^{-\tau/T} \hspace{0.1cm}{\rm d}\tau .$$ | \frac{1}{T^2} \cdot \int_{ 0 }^{ t } \tau \cdot {\rm e}^{-\tau/T} \hspace{0.1cm}{\rm d}\tau .$$ | ||
− | + | *With the substitution $u = τ/T$ and using the given integral it follows that: | |
− | $$ | + | :$${\rm \sigma_2}(t) = \int_{ 0 }^{ t /T} u \cdot {\rm e}^{-u} \hspace{0.1cm}{\rm d}u ={\rm e}^{-u} \cdot (-u-1) |_{ 0 }^{ t /T} \hspace{0.5cm} \Rightarrow \hspace{0.5cm}{\rm \sigma_2}(t) = |
− | 1- \left( 1 + {t}/{T} \right) \cdot {\rm e}^{-t/T}. | + | 1- \left( 1 + {t}/{T} \right) \cdot {\rm e}^{-t/T}.$$ |
− | + | *The following is obtained at the given times taking the factor $2\ \rm V$ further into account: | |
− | $$ | + | :$$y_2(t = T) = {\rm 2 \,V} \cdot \left( 1- 2 \cdot {\rm e}^{-1} \right) \hspace{0.15cm}\underline{= {\rm 0.528 \,V}}, \hspace{0.9cm}y_2(t = 5T) = {\rm 2 \,V} \cdot \left( 1- 6 \cdot {\rm e}^{-5} \right) \hspace{0.15cm}\underline{= {\rm 1.919 \,V}}.$$ |
− | + | *For even larger times $y_2(t)$ approaches more and more the final value $2\hspace{0.05cm} \rm V$ . | |
{{ML-Fuß}} | {{ML-Fuß}} | ||
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− | [[Category: | + | [[Category:Linear and Time-Invariant Systems: Exercises|^1.2 System Description in Time Domain^]] |
Latest revision as of 01:51, 19 July 2021
In Exercise 1.1 and Exercise 1.1Z of the chapter System Description in Frequency Domain the so-called "RC low-pass filters" were described in the frequency domain. Now, the time domain representation is elaborated on.
The circuit with the input signal $x(t)$ and the output signal $y_1(t)$ is a low-pass filter of first order with frequency response
- $$H_{\rm 1}(f) = \frac{1}{1+{\rm j}\cdot f/f_{\rm 0}}.$$
Here, $f_0 = 1/(2πRC)$ indicates the 3dB cut-off frequency. If a Dirac-shaped signal $x(t) = δ(t)$ is applied to the input, the signal $y_1(t)$ is given by the system output according to the sketch in the middle.
The relationship between the system parameters $R$ (resistance), $C$ (capacitance) and $T$ is (see Exercise 1.3Z):
- $$T = \frac{1}{\omega_{\rm 0}}= \frac{1}{2\pi f_{\rm 0}} = R \cdot C.$$
For numerical calculations $T = 1 \ \rm ms$ shall be used in the following.
The circuit below with input $x(t)$ and output $y_2(t)$ represents a low-pass filter of second order:
- $$H_{\rm 2}(f) = \big[H_{\rm 1}(f)\big]^2 =\frac{1}{(1+{\rm j}\cdot f/f_{\rm 0})^2}.$$
- The impulse response associated with $H_2(f)$ is $h_2(t)$.
- The system parameter $f_0$ no longer indicates the 3dB cut-off frequency for a low-pass filter of second or higher order.
- Furthermore, it should be noted that the two RC elements must be decoupled in order to achieve impedance matching.
- An operational amplifier, for example, is suitable for this. However, this hint is not relevant for the solution of this task.
Please note:
- The exercise belongs to the chapter System Description in Time Domain.
- The following indefinite integral is given:
- $$\int u \cdot {\rm e}^{a \cdot \hspace{0.03cm} u} \hspace{0.1cm}{\rm d}u = \frac{{\rm e}^{\hspace{0.03cm}a \cdot \hspace{0.03cm} u}}{a^2} \cdot (a \cdot u -1).$$
Questions
Solution
- According to the solution of Exercise 1.3Z and the sketch above the following holds:
- $$h_1(t) = y_1(t) ={1}/{T} \cdot {\rm e}^{-t/T}.$$
- For the time $t_1$ the following shall hold:
- $$h_1(t_{\rm 1}) ={1}/{T} \cdot {\rm e}^{-t_{\rm 1}/T} = \frac{1}{2T} \hspace{0.5cm}\Rightarrow\hspace{0.5cm}{t_{\rm 1}}/{T} = {\rm ln}(2)\hspace{0.1cm} \Rightarrow \hspace{0.1cm}t_{\rm 1} = 0.693 \cdot T \hspace{0.15cm}\underline{= {\rm 0.693\,\,ms}}.$$
(2) The input signal $x(t)$ is an exponentially decreasing impulse like the impulse response $h_1(t)$ but dimensionless.
- Thus, according to the convolution theorem:
- $$y_1(t) = x (t) * h_1 (t) = T \cdot \big[ h_1 (t) * h_1 (t) \big].$$
- The convolution is illustrated here for a specific time $t$ by a sketch.
- The following is obtained after renaming the variables:
- $$\begin{align*}h_1(\tau) & = \frac{1}{T} \cdot {\rm e}^{-\tau/T}\hspace{0.1cm} \Rightarrow \hspace{0.1cm}h_1(-\tau) = \frac{1}{T} \cdot {\rm e}^{\tau/T}\hspace{0.1cm} \Rightarrow \hspace{0.1cm}h_1(t-\tau) = \frac{1}{T} \cdot {\rm e}^{(-t+\tau)/T} \\ & \Rightarrow\hspace{0.5cm} y_1(t) = T \cdot \int\limits_{ - \infty }^{ + \infty } h_1 ( {t - \tau } )\cdot {h_1 ( \tau )} \hspace{0.1cm}{\rm d}\tau.\end{align*}$$
- For $τ < 0$ the impulse response is $h_1(τ) = 0$. For $τ > t$ the first convolution operand vanishes (see sketch). From this it follows that:
- $$y_1(t) = T \cdot \frac{1}{T^2}\cdot \int_{ 0 }^{ t } {\rm e}^{(-t+\tau)/T} \cdot {\rm e}^{-\tau/T} \hspace{0.1cm}{\rm d}\tau = \frac{1}{T}\cdot \int_{ 0 }^{ t } {\rm e}^{-t/T} \hspace{0.1cm}{\rm d}\tau .$$
- Now, it can be observed that the integrand is independent of the integration variable $τ$. Consequently, the following holds:
- $$y_1(t) = {t}/{T} \cdot {\rm e}^{-t/T}\hspace{0.1cm} \Rightarrow \hspace{0.1cm} y_1(t =0) \hspace{0.15cm}\underline{=0}; \hspace{0.5cm}y_1(t =T)={\rm e}^{-1} \hspace{0.15cm}\underline{=0.368}.$$
(3) Due to the interrelationship $H_2(f) = H_1(f) · H_1(f)$ the following holds for the impulse response: $h_2(t) = h_1 (t) * h_1 (t).$
- Except for the additional constant factor $1/T$ the same result as in subtask (2) is obtained:
- $$h_2(t) ={t}/{T^2} \cdot {\rm e}^{-t/T}.$$
- The maximum value is obtained by setting the derivative to zero:
- $$\frac{{\rm d} h_2(t)}{{\rm d}t} = \frac{1}{T^2} \cdot {\rm e}^{-t/T} \cdot \left( 1 - {t}/{T}\right) = 0 \hspace{0.5cm} \Rightarrow \hspace{0.5cm} t_{\rm 2} \hspace{0.15cm}\underline{= T = 1\,\,{\rm ms}} \hspace{0.5cm}\Rightarrow \hspace{0.5cm} h_2(t_{\rm 2}) =\frac{{\rm e}^{-1}}{T} =\frac{0.368}{1\,\,{\rm ms}} \hspace{0.15cm}\underline{ = {\rm 368 \hspace{0.1cm} 1/s}}.$$
(4) In general or with the result from (3) the following applies to the step response:
- $${\rm \sigma_2}(t) = \int_{ 0 }^{ t } {h_2 ( \tau )} \hspace{0.1cm}{\rm d}\tau = \frac{1}{T^2} \cdot \int_{ 0 }^{ t } \tau \cdot {\rm e}^{-\tau/T} \hspace{0.1cm}{\rm d}\tau .$$
- With the substitution $u = τ/T$ and using the given integral it follows that:
- $${\rm \sigma_2}(t) = \int_{ 0 }^{ t /T} u \cdot {\rm e}^{-u} \hspace{0.1cm}{\rm d}u ={\rm e}^{-u} \cdot (-u-1) |_{ 0 }^{ t /T} \hspace{0.5cm} \Rightarrow \hspace{0.5cm}{\rm \sigma_2}(t) = 1- \left( 1 + {t}/{T} \right) \cdot {\rm e}^{-t/T}.$$
- The following is obtained at the given times taking the factor $2\ \rm V$ further into account:
- $$y_2(t = T) = {\rm 2 \,V} \cdot \left( 1- 2 \cdot {\rm e}^{-1} \right) \hspace{0.15cm}\underline{= {\rm 0.528 \,V}}, \hspace{0.9cm}y_2(t = 5T) = {\rm 2 \,V} \cdot \left( 1- 6 \cdot {\rm e}^{-5} \right) \hspace{0.15cm}\underline{= {\rm 1.919 \,V}}.$$
- For even larger times $y_2(t)$ approaches more and more the final value $2\hspace{0.05cm} \rm V$ .