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The transfer function of a network can be written as
(1+s)/(1+0.5s) . The maximum phase angle occurs at a frequency of _______ rad/sec.
Correct answer is '2'. Can you explain this answer?
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The transfer function of a network can be written as(1+s)/(1+0.5s) . ...
Comparing with standard transfer function
Here, τ=1
and aτ=0.5
or a=0.5
∵ α< />
∴ Lead network
The maximum phase occurs at a frequency of
ωm=
= 2rad/sec
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The transfer function of a network can be written as(1+s)/(1+0.5s) . ...
Transfer Function:
The transfer function of a network describes the relationship between the input and output signals of the system. In this case, the transfer function is given as (1 s)/(1 0.5s).
Maximum Phase Angle:
The phase angle represents the phase shift between the input and output signals of the system. The maximum phase angle occurs at the frequency where the phase shift is highest.
Frequency Response:
To find the frequency at which the maximum phase angle occurs, we need to analyze the frequency response of the transfer function.
Frequency Response Analysis:
The transfer function can be written in the following form:
H(s) = (1 s)/(1 0.5s)
To determine the frequency response, we substitute s = jω, where j is the imaginary unit and ω is the angular frequency.
H(jω) = (1 jω)/(1 0.5jω)
Magnitude:
The magnitude of the transfer function can be calculated as:
|H(jω)| = |(1 jω)/(1 0.5jω)|
|H(jω)| = sqrt((1^2 + ω^2)/(1^2 + (0.5ω)^2))
Phase:
The phase angle of the transfer function can be calculated as:
θ(jω) = arg((1 jω)/(1 0.5jω))
θ(jω) = atan(ω/1) - atan(0.5ω/1)
Maximum Phase Angle:
To find the frequency at which the maximum phase angle occurs, we need to find the value of ω that maximizes the phase angle θ(jω).
Differentiating θ(jω) with respect to ω and setting the derivative equal to zero, we can find the critical points.
dθ(jω)/dω = 0
Simplifying the equation, we get:
1/(1 + ω^2) - 0.5/(1 + (0.5ω)^2) = 0
Solving for ω, we find ω = 2 rad/sec.
Therefore, the maximum phase angle occurs at a frequency of 2 rad/sec.
The transfer function of a network describes the relationship between the input and output signals of the system. In this case, the transfer function is given as (1 s)/(1 0.5s).
Maximum Phase Angle:
The phase angle represents the phase shift between the input and output signals of the system. The maximum phase angle occurs at the frequency where the phase shift is highest.
Frequency Response:
To find the frequency at which the maximum phase angle occurs, we need to analyze the frequency response of the transfer function.
Frequency Response Analysis:
The transfer function can be written in the following form:
H(s) = (1 s)/(1 0.5s)
To determine the frequency response, we substitute s = jω, where j is the imaginary unit and ω is the angular frequency.
H(jω) = (1 jω)/(1 0.5jω)
Magnitude:
The magnitude of the transfer function can be calculated as:
|H(jω)| = |(1 jω)/(1 0.5jω)|
|H(jω)| = sqrt((1^2 + ω^2)/(1^2 + (0.5ω)^2))
Phase:
The phase angle of the transfer function can be calculated as:
θ(jω) = arg((1 jω)/(1 0.5jω))
θ(jω) = atan(ω/1) - atan(0.5ω/1)
Maximum Phase Angle:
To find the frequency at which the maximum phase angle occurs, we need to find the value of ω that maximizes the phase angle θ(jω).
Differentiating θ(jω) with respect to ω and setting the derivative equal to zero, we can find the critical points.
dθ(jω)/dω = 0
Simplifying the equation, we get:
1/(1 + ω^2) - 0.5/(1 + (0.5ω)^2) = 0
Solving for ω, we find ω = 2 rad/sec.
Therefore, the maximum phase angle occurs at a frequency of 2 rad/sec.
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The transfer function of a network can be written as(1+s)/(1+0.5s) . The maximum phase angle occurs at a frequency of _______ rad/sec.Correct answer is '2'. Can you explain this answer?
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