Deriving M_r, ω_r and BW

1 · Substitute s = iω and find the magnitude

First use (iω)² = −ω² to separate the real and imaginary parts of the denominator.

G(iω) = ω_n²(iω)² + 2ζω_n(iω) + ω_n²

= ω_n²(ω_n² − ω²) + i(2ζω_nω)

For a complex number a + ib, its magnitude is √(a² + b²). Divide the numerator magnitude by the denominator magnitude:

|G(iω)| = ω_n²√[(ω_n² − ω²)² + (2ζω_nω)²]

2 · Normalize the frequency

Let x = ω/ω_n, so ω = ω_nx. Substitute this into both terms:

ω_n² − ω² = ω_n² − ω_n²x² = ω_n²(1 − x²)

2ζω_nω = 2ζω_n(ω_nx) = ω_n²(2ζx)

After squaring, both denominator terms contain ω_n⁴. Take ω_n² outside the square root and cancel it:

|G(iω)| = ω_n²ω_n²√[(1 − x²)² + 4ζ²x²]

= 1√[(1 − x²)² + 4ζ²x²]

Call the expression inside the square root D(x):

D(x) = (1 − x²)² + 4ζ²x²

= 1 − 2x² + x⁴ + 4ζ²x²

= x⁴ + (4ζ² − 2)x² + 1

Magnitude = 1/√D(x). A smaller D gives a larger magnitude.

3 · Find the resonant frequency ω_r

To locate the largest magnitude, find the smallest D(x). Differentiate with respect to x:

D′(x) = 4x³ + 2(4ζ² − 2)x

= 4x³ + (8ζ² − 4)x

= 4x(x² + 2ζ² − 1)

Set the derivative to zero:

4x(x² + 2ζ² − 1) = 0

x = 0   or   x² = 1 − 2ζ²

A positive-frequency solution exists when 1 − 2ζ² > 0, or 0 < ζ < 1/√2.

x_r = √(1 − 2ζ²)

Before x_r, D′ is negative; after x_r, it is positive. D therefore reaches a minimum at x_r, and the magnitude reaches a maximum.

Finally convert back from x to ω using ω = ω_nx:

ω_r = ω_n√(1 − 2ζ²)

4 · Substitute that frequency to find M_r

At the peak, x_r² = 1 − 2ζ², so:

1 − x_r² = 1 − (1 − 2ζ²) = 2ζ²

Now evaluate the denominator at x_r:

D(x_r) = (1 − x_r²)² + 4ζ²x_r²

= (2ζ²)² + 4ζ²(1 − 2ζ²)

= 4ζ⁴ + 4ζ² − 8ζ⁴

= 4ζ² − 4ζ⁴ = 4ζ²(1 − ζ²)

The peak magnitude is 1/√D(x_r). Since ζ > 0, √(4ζ²) = 2ζ:

M_r = 1√[4ζ²(1 − ζ²)] = 12ζ√(1 − ζ²)

5 · Why does the table have two damping ranges?

When ζ ≥ 1/√2, the factor x² + 2ζ² − 1 is positive for every x > 0. Therefore D′(x) > 0: D increases, and the magnitude decreases.

The maximum occurs at x = 0, which means ω = 0:

D(0) = 1   ⇒   |G(0)| = 1

M_r = 1,   ω_r = 0

There is no nonzero-frequency resonant peak in this range. At ζ = 1/√2, the peak formulas also give M_r = 1 and ω_r = 0.

6 · Use the bandwidth condition

Bandwidth uses the low-frequency magnitude as its reference. For this system:

G(0) = ω_n²ω_n² = 1

Define x_B = BW/ω_n. At bandwidth:

|G(iBW)| = |G(0)|√2 = 1√2 ≈ 0.7071

Substitute |G| = 1/√D and square both sides:

1√D(x_B) = 1√2   ⇒   1D(x_B) = 12

D(x_B) = 2

Insert the expanded expression for D and subtract 2:

x_B⁴ + (4ζ² − 2)x_B² + 1 = 2

x_B⁴ + (4ζ² − 2)x_B² − 1 = 0

Let u = x_B², so x_B⁴ = u². This becomes a quadratic:

u² + (4ζ² − 2)u − 1 = 0

7 · Solve the quadratic and convert back to frequency

For au² + bu + c = 0, use u = [−b ± √(b² − 4ac)]/(2a). Here a = 1, b = 4ζ² − 2, and c = −1:

u = −(4ζ² − 2) ± √[(4ζ² − 2)² + 4]2

Expand and simplify the expression inside the square root:

(4ζ² − 2)² + 4 = 16ζ⁴ − 16ζ² + 4 + 4

= 16ζ⁴ − 16ζ² + 8

= 4(4ζ⁴ − 4ζ² + 2)

Take √4 = 2 outside the square root:

u = 2 − 4ζ² ± 2√(4ζ⁴ − 4ζ² + 2)2

= 1 − 2ζ² ± √(4ζ⁴ − 4ζ² + 2)

The minus root is negative: √[(4ζ² − 2)² + 4] > |4ζ² − 2|. Since u = x_B² ≥ 0, keep the plus root:

u = 1 − 2ζ² + √(4ζ⁴ − 4ζ² + 2)

Finally, x_B = √u and BW = ω_nx_B:

BW = ω_n [(1 − 2ζ²) + √(4ζ⁴ − 4ζ² + 2)]^1/2

M_r and ω_r: minimize D(x).   BW: solve D(x_B) = 2.