Bandpass Function, Center Frequency, Half Power Frequency, Bandwidth, Quality Factor And Resonance
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Expressions Of Pj Problems
Bandpass Function, Center Frequency, Half Power Frequency, Bandwidth, Quality Factor And Resonance

(a) Indicate the general representation of the bandpass function in the s domain
(b) Indicate the frequency response of the bandpass function
(c) Determine the center frequency of the frequency response
(d) Determine the half-power frequencies of the frequency response
(e) Determine the bandwidth of the frequency response
(f) Determine the quality factor
(g) Relate the quality factor to resonance
(h) If H(s) = s/(s2 + as + b), determine a and b such that the magnitude of the frequency response |H(ω)| has a maximum at 100 Hz with a half-power bandwidth of 5 Hz. What is the quality factor?

The strings: S7P5A51 (change - physical).

The math:
Pj Problem of Interest is of type change (change - physical).

(a) General bandpass function in the s domain:
H(s) = ks/(s2 + as + b)------(1) Where k, a, b > 0.

(b) Frequency response of bandpass function:
H(jω) = k/(2 + a + b) = k/(b - ω2 + ajω)------(2)

(c) |H(jω)|2 = k2ω2/[(b - ω2)2 + a2ω2)] = k2/[a2 + (b - ω2)22]------(3)
The frequency (ωo) at which |H| is maximum is the center frequency.
So from eq(3):|H| is maximum when (b - ω2) = 0 that is, when ω = √b
So, center frequency, ωo = √b.
At the center frequency, |H|max = k/a.

(d) Consider the following equalities:
|Hl|2 = |Hh|2 = (1/2)|Ho)|2-------(4)
ωl and ωh in eq(4) are called the half-power lower frequency and the half-power upper frequency respectively.
Substituting eq(3) into eq(4) and solving, we have:
ωl and ωh as the roots of:
(b - ω2)22] = a2
So, ωl = √(a2/4 + b) - a/2
And ωh = √(a2/4 + b) + a/2

(e) Bandwidth, β = ωh - ωl =√(a2/4 + b) + a/2 - √(a2/4 + b) - a/2 = a.
ωhωl = b = (ωo)2

(f) Quality factor, Q = ωo/β = √b/a
Quality factor measures sharpness of frequency response around the center frequency.

(g) Increasing quality factor tends towards resonance.

(h) Bandwidth = β = a = 2π(5) = 31.416
ωo = 2π(100) = 628.3185 = √b
So, b = (628.3185)2 = 394784
Quality factor, Q = ωo/β = 628.3185/31.416 = 20


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