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And

arg ( H ( ) ) = arg ( K ) + arg ( j ωτ z1 + 1 ) + arg ( j ωτ z2 + 1 ) + . . . + arg ( j ωτ zM + 1 ) arg ( j ωτ p1 + 1 ) arg ( j ωτ p2 + 1 ) . . . arg ( j ωτ pN + 1 ) alignl { stack { size 12{"arg" \( H \( jω \) \) ="arg" \( K \) +"arg" \( j ital "ωτ" rSub { size 8{z1} } +1 \) +"arg" \( j ital "ωτ" rSub { size 8{z2} } +1 \) + "." "." "." +"arg" \( j ital "ωτ" rSub { size 8{ ital "zM"} } +1 \) } {} #matrix { matrix {matrix { matrix {{} # {} } {} # {}} {} # {} } {} # {}} - "arg" \( j ital "ωτ" rSub { size 8{p1} } +1 \) - "arg" \( j ital "ωτ" rSub { size 8{p2} } +1 \) - "." "." "." - "arg" \( j ital "ωτ" rSub { size 8{ ital "pN"} } +1 \) {} } } {}

4/ Logarithmic magnitude

Taking twenty times the logarithm of the magnitude yields

20 log 10 H ( ) = 20 log 10 K + 20 log 10 j ωτ z1 + 1 + 20 log 10 j ωτ z2 + 1 + . . . + 20 log 10 j ωτ zM + 1 20 log 10 j ωτ p1 + 1 20 log 10 j ωτ p2 + 1 . . . 20 log 10 j ωτ pN + 1 alignl { stack { size 12{"20""log" rSub { size 8{"10"} } \lline H \( jω \) \lline ="20""log" rSub { size 8{"10"} } \lline K \lline +"20""log" rSub { size 8{"10"} } \lline j ital "ωτ" rSub { size 8{z1} } +1 \lline +"20""log" rSub { size 8{"10"} } \lline j ital "ωτ" rSub { size 8{z2} } +1 \lline + "." "." "." +"20""log" rSub { size 8{"10"} } \lline j ital "ωτ" rSub { size 8{ ital "zM"} } +1 \lline } {} #matrix { matrix {matrix { matrix {{} # {} } {} # {}} {} # {} } {} # {}} - "20""log" rSub { size 8{"10"} } \lline j ital "ωτ" rSub { size 8{p1} } +1 \lline - "20""log" rSub { size 8{"10"} } \lline j ital "ωτ" rSub { size 8{p2} } +1 \lline - "." "." "." - "20""log" rSub { size 8{"10"} } \lline j ital "ωτ" rSub { size 8{ ital "pN"} } +1 \lline {}} } {}

Note than both the logarithmic magnitude and the angle are expressed as sums of terms of the form

± 20 log 10 j ωτ + 1 and ± arg ( j ωτ + 1 ) size 12{ +- "20 log" rSub { size 8{"10"} } \lline j ital "ωτ" +1 \lline " " matrix { {} # {}} " and " matrix { {} # {}} +- " arg" \( j ital "ωτ" +1 \) } {}

Therefore, to plot the frequency response we need to add terms of the above form.

5/ Decibels

It is common to plot frequency responses as Bode diagrams whose magnitude is expressed in decibels. The decibel, denoted by dB, is defined as 20 log 10 Η size 12{"20 log" rSub { size 8{"10"} } \lline Η \lline } {} . The following table gives decibel equivalents for a few quantities.

How many decibels correspond to |H| = 50? Express |H| =100/2. Then

20 log 10 ( 100/2 ) = 20 log 10 100 - 20log 10 2 40 - 6 = 34 dB size 12{"20 log" rSub { size 8{"10"} } \( "100/2" \) =" 20 log" rSub { size 8{"10"} } " 100 - 20log" rSub { size 8{"10"} } " 2" approx " 40" "- 6"" = ""34" "dB"} {}

6/ Asymptotes

To plot the frequency response of a system with real poles and zeros, we need to plot terms of the form

± 20 log 10 1 + j ωτ and ± arg ( 1 + j ωτ ) size 12{ +- "20 log" rSub { size 8{"10"} } \lline 1+j ital "ωτ" \lline " " matrix { {} # {}} " and " matrix { {} # {}} " " +- " arg" \( 1+j ital "ωτ" \) } {}

The low and high frequency asymptotes are

7/ Corner frequency

At ω = ω c = 1 / τ size 12{ω "= "ω rSub { size 8{c} } " = 1"/τ} {} , called the corner or cut-off frequency,

  • the low- and high-frequency asymptotes intersect,
  • the magnitude is

± 20 log 10 1+j ωτ = ± 20 log 10 1+j1 = ± 20 log 10 2 1/2 ± 3 dB size 12{ +- "20 log" rSub { size 8{"10"} } \lline "1+j" ital "ωτ" \lline " = " +- "20 log" rSub { size 8{"10"} } \lline "1+j1" \lline " = " +- "20 log" rSub { size 8{"10"} } 2 rSup { size 8{"1/2"} } " " +- "3 dB"} {}

  • the angle is

± arg ( 1+ j ωτ ) = ± arg ( 1+ j ) = ± π / 4 size 12{ +- "arg" \( "1+"j ital "ωτ" \) " = " +- "arg" \( "1+"j \) " = " +- π/4} {}

Example — first-order lowpass system

First-order low pass systems arise in a large variety of physical contexts. For example,

For the parameters M = B = R = C = 1, the frequency responses for the two systems are

H ( ) = V ( ) F ( ) = 1 + 1 and H ( ) = V 0 ( ) V i ( ) = 1 + 1 size 12{H \( jω \) = { {V \( jω \) } over {F \( jω \) } } = { {1} over {jω+1} } matrix { {} # {}} ital "and" matrix { {} # {}} H \( jω \) = { {V rSub { size 8{0} } \( jω \) } over {V rSub { size 8{i} } \( jω \) } } = { {1} over {jω+1} } } {}

Magnitude

For

H ( ) = 1 + 1 size 12{H \( jω \) = { {1} over {jω+1} } } {}

the low-frequency asymptote has a slope of 0 and an intercept of 0 dB and the high-frequency asymptote has a slope of -20 dB/decade and an intercept of 0 dB at the corner frequency.

The corner frequency is 1 rad/sec and the bandwidth is 1 rad/sec. The two asymptotes intersect at ω = 1 where 20 log 10 Η ( ) = -3 dB size 12{"20 log" rSub { size 8{"10"} } \lline Η \( "jω" \) \lline "= ""-3" ital "dB"} {}

H ( ) = 1 + 1 size 12{H \( jω \) = { {1} over {jω+1} } } {}

the low- and high-frequency asymptotes of the angle of the frequency response are 0 and 90 0 size 12{-"90" rSup { size 8{0} } } {} (−π/2 radians), respectively. The angle is 45 0 size 12{-"45" rSup { size 8{0} } } {} at the corner frequency (1 rad/sec).

A line drawn from the low frequency asymptote a decade below the corner frequency to the high frequency asymptote a decade above the corner frequency approximates the angle of the frequency response.

Physical interpretation

With M = B = 1, the frequency response is

H ( ) = V ( ) F ( ) = 1 + 1 size 12{H \( jω \) = { {V \( jω \) } over {F \( jω \) } } = { {1} over {jω+1} } } {}

At low frequencies, |H(jω)| → 1 and arg H(jω) → 0. The inertia of the mass is negligible, and the damping force dominates so that the external force is proportional to velocity.

At high frequencies, |H(jω)| → 1/ω and arg H(jω) → 90 0 size 12{-"90" rSup { size 8{0} } } {} . The inertia of the mass dominates so that the acceleration is proportional to external force and the velocity decreases as frequency increases.

With R = C = 1, the frequency response is

{} H ( ) = V o ( ) V i ( ) = 1 + 1 size 12{H \( jω \) = { {V rSub { size 8{o} } \( jω \) } over {V rSub { size 8{i} } \( jω \) } } = { {1} over {jω+1} } } {}

At low frequencies, |H(jω)| → 1 and arg H(jω) → 0. The impedance of the capacitance is large so that all the input voltage appears at the output.

At high frequencies, |H(jω)| → 1/ω and arg H(jω) → 90 0 size 12{-"90" rSup { size 8{0} } } {} . The impedance of the capacitance is small so that the current is determined by the resistance and the output voltage is determined by the impedance of the capacitance which decreases as frequency increases.

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Source:  OpenStax, Signals and systems. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10803/1.1
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