Difference between revisions of "Aufgaben:Exercise 1.4: Low-Pass Filter of 2nd Order"
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[[File:EN_LZI_A_1_4.png|right|frame|Low-pass filter of first order (top) <br>and of second order (bottom)]] | [[File:EN_LZI_A_1_4.png|right|frame|Low-pass filter of first order (top) <br>and of second order (bottom)]] | ||
− | In [[Aufgaben:Exercise_1.1:_Simple_Filter_Functions| | + | 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 | + | 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 | + | 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 $R$, $C$ and $T$ is (see [[Aufgaben:Exercise_1.3Z:_Exponentially_Decreasing_Impulse_Response| | + | 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]]): |
:$$T = \frac{1}{\omega_{\rm 0}}= \frac{1}{2\pi f_{\rm 0}} = R \cdot C.$$ | :$$T = \frac{1}{\omega_{\rm 0}}= \frac{1}{2\pi f_{\rm 0}} = R \cdot C.$$ | ||
For numerical calculations $T = 1 \ \rm ms$ shall be used in the following. | 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: | The circuit below with input $x(t)$ and output $y_2(t)$ represents a low-pass filter of second order: | ||
:$$H_{\rm 2}(f) = \big[H_{\rm 1}(f)\big]^2 =\frac{1}{(1+{\rm j}\cdot f/f_{\rm 0})^2}.$$ | :$$H_{\rm 2}(f) = \big[H_{\rm 1}(f)\big]^2 =\frac{1}{(1+{\rm j}\cdot f/f_{\rm 0})^2}.$$ | ||
*The impulse response associated with $H_2(f)$ is $h_2(t)$. | *The impulse response associated with $H_2(f)$ is $h_2(t)$. | ||
− | *The system parameter $f_0$ no longer indicates the | + | *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. | *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. | + | *An operational amplifier, for example, is suitable for this. However, this hint is not relevant for the solution of this task. |
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''Please note:'' | ''Please note:'' | ||
− | *The | + | *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: | *The following indefinite integral is given: | ||
:$$\int u \cdot {\rm e}^{a \cdot \hspace{0.03cm} u} \hspace{0.1cm}{\rm d}u = | :$$\int u \cdot {\rm e}^{a \cdot \hspace{0.03cm} u} \hspace{0.1cm}{\rm d}u = | ||
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− | {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 $t = T$ | + | {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% } $\rm V$ | $y_2(t = T) \ = \ $ { 0.528 5% } $\rm V$ | ||
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</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
'''(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. | '''(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. | ||
− | * | + | *According to the [[Aufgaben:Exercise_1.3Z:_Exponentially_Decreasing_Impulse_Response|solution of Exercise 1.3Z]] and the sketch above the following holds: |
:$$h_1(t) = y_1(t) ={1}/{T} \cdot {\rm e}^{-t/T}.$$ | :$$h_1(t) = y_1(t) ={1}/{T} \cdot {\rm e}^{-t/T}.$$ | ||
− | * | + | *For the time $t_1$ the following shall hold: |
:$$h_1(t_{\rm 1}) ={1}/{T} \cdot {\rm e}^{-t_{\rm 1}/T} = \frac{1}{2T} | :$$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}}.$$ | \hspace{0.5cm}\Rightarrow\hspace{0.5cm}{t_{\rm 1}}/{T} = {\rm ln}(2)\hspace{0.1cm} \Rightarrow \hspace{0.1cm}t_{\rm 1} = 0.693 \cdot T \hspace{0.15cm}\underline{= {\rm 0.693\,\,ms}}.$$ | ||
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− | '''(2)''' | + | '''(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| | + | [[File:P_ID830__LZI_A_1_4b.png |right|frame| Illustration of the convolution]] |
− | * | + | *Thus, according to the convolution theorem: |
:$$y_1(t) = x (t) * h_1 (t) = T \cdot \big[ h_1 (t) * h_1 (t) \big].$$ | :$$y_1(t) = x (t) * h_1 (t) = T \cdot \big[ h_1 (t) * h_1 (t) \big].$$ | ||
− | * | + | *The convolution is illustrated here for a specific time $t$ by a sketch. |
− | * | + | *The following is obtained after renaming the variables: |
:$$\begin{align*}h_1(\tau) & = \frac{1}{T} \cdot {\rm e}^{-\tau/T}\hspace{0.1cm} \Rightarrow \hspace{0.1cm}h_1(-\tau) = \frac{1}{T} \cdot {\rm e}^{\tau/T}\hspace{0.1cm} \Rightarrow \hspace{0.1cm}h_1(t-\tau) = \frac{1}{T} \cdot {\rm e}^{(-t+\tau)/T} \\ & \Rightarrow\hspace{0.5cm} | :$$\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: |
:$$y_1(t) = {t}/{T} \cdot {\rm e}^{-t/T}\hspace{0.1cm} \Rightarrow | :$$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}.$$ | \hspace{0.1cm} y_1(t =0) \hspace{0.15cm}\underline{=0}; \hspace{0.5cm}y_1(t =T)={\rm e}^{-1} \hspace{0.15cm}\underline{=0.368}.$$ | ||
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− | '''(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: |
:$$h_2(t) ={t}/{T^2} \cdot {\rm e}^{-t/T}.$$ | :$$h_2(t) ={t}/{T^2} \cdot {\rm e}^{-t/T}.$$ | ||
− | * | + | *The maximum value is obtained by setting the derivative to zero: |
:$$\frac{{\rm d} h_2(t)}{{\rm d}t} = \frac{1}{T^2} \cdot {\rm e}^{-t/T} \cdot \left( 1 - {t}/{T}\right) = 0 \hspace{0.5cm} \Rightarrow \hspace{0.5cm} t_{\rm 2} \hspace{0.15cm}\underline{= T = 1\,\,{\rm ms}} \hspace{0.5cm}\Rightarrow \hspace{0.5cm} h_2(t_{\rm 2}) =\frac{{\rm e}^{-1}}{T} =\frac{0.368}{1\,\,{\rm ms}} \hspace{0.15cm}\underline{ = {\rm 368 \hspace{0.1cm} 1/s}}.$$ | :$$\frac{{\rm d} h_2(t)}{{\rm d}t} = \frac{1}{T^2} \cdot {\rm e}^{-t/T} \cdot \left( 1 - {t}/{T}\right) = 0 \hspace{0.5cm} \Rightarrow \hspace{0.5cm} t_{\rm 2} \hspace{0.15cm}\underline{= T = 1\,\,{\rm ms}} \hspace{0.5cm}\Rightarrow \hspace{0.5cm} h_2(t_{\rm 2}) =\frac{{\rm e}^{-1}}{T} =\frac{0.368}{1\,\,{\rm ms}} \hspace{0.15cm}\underline{ = {\rm 368 \hspace{0.1cm} 1/s}}.$$ | ||
− | '''(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: |
:$${\rm \sigma_2}(t) = \int_{ 0 }^{ t /T} u \cdot {\rm e}^{-u} \hspace{0.1cm}{\rm d}u ={\rm e}^{-u} \cdot (-u-1) |_{ 0 }^{ t /T} \hspace{0.5cm} \Rightarrow \hspace{0.5cm}{\rm \sigma_2}(t) = | :$${\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: |
:$$y_2(t = T) = {\rm 2 \,V} \cdot \left( 1- 2 \cdot {\rm e}^{-1} \right) \hspace{0.15cm}\underline{= {\rm 0.528 \,V}}, \hspace{0.9cm}y_2(t = 5T) = {\rm 2 \,V} \cdot \left( 1- 6 \cdot {\rm e}^{-5} \right) \hspace{0.15cm}\underline{= {\rm 1.919 \,V}}.$$ | :$$y_2(t = T) = {\rm 2 \,V} \cdot \left( 1- 2 \cdot {\rm e}^{-1} \right) \hspace{0.15cm}\underline{= {\rm 0.528 \,V}}, \hspace{0.9cm}y_2(t = 5T) = {\rm 2 \,V} \cdot \left( 1- 6 \cdot {\rm e}^{-5} \right) \hspace{0.15cm}\underline{= {\rm 1.919 \,V}}.$$ | ||
− | * | + | *For even larger times $y_2(t)$ approaches more and more the final value $2\hspace{0.05cm} \rm V$ . |
{{ML-Fuß}} | {{ML-Fuß}} |
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 $2\hspace{0.05cm} \rm V$ .