Trapezoidal Rule Stability

Calculus Level 3

Consider the formula for trapezoidal integration of a complex exponential.

y ( t ) = y ( t − Δ t ) + 1 2 [ y ˙ ( t ) + y ˙ ( t − Δ t ) ] Δ t y ( t ) = e λ t y ˙ ( t ) = λ y ( t ) λ = a + j b j = − 1 y(t) = y(t - \Delta t) + \frac{1}{2} \Big [ \dot{y}(t) + \dot{y}(t - \Delta t) \Big ] \, \Delta t \\ y(t) = e^{\lambda \, t} \\ \dot{y}(t) = \lambda \, y(t) \\ \lambda = a + j \, b \\ j = \sqrt{-1}

The complex exponential is processed sequentially in time with a discrete time interval Δ t \Delta t . In order for the algorithm to be stable (non-divergent) for a given Δ t \Delta t , the following condition must hold true:

∣ y ( t ) y ( t − Δ t ) ∣ ≤ 1 \large{\Big| \frac{y(t)}{y(t - \Delta t)} \Big | \leq 1}

Further, the algorithm "converges" ( ∣ y ( ∞ ) ∣ ≈ 0 ) (|y(\infty)| \approx 0) if the ratio has a magnitude less than 1 1 .

Suppose Δ t = 1 \Delta t = 1 . In terms of the ( a , b ) (a,b) plane, which of the given options best describes the trapezoidal rule "convergence region".

Full left-hand plane ( a < 0 ) (a < 0) , plus a finite area within ( a > 0 ) (a > 0) region Only a finite area within the ( a < 0 ) (a < 0) region (with the rest of the complex plane excluded) Full left-hand plane ( a < 0 ) (a < 0) , and nothing within the ( a > 0 ) (a > 0) region

Steven Chase by Steven Chase , 37, USA

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1 solution

Steven Chase
May 28, 2019

Find the ratio of y ( t ) y(t) to y ( t − Δ t ) y(t - \Delta t)

y ( t ) = y ( t − Δ t ) + 1 2 [ y ˙ ( t ) + y ˙ ( t − Δ t ) ] Δ t y ( t ) = y ( t − Δ t ) + 1 2 [ λ y ( t ) + λ y ( t − Δ t ) ] Δ t y ( t ) [ 1 − 1 2 λ Δ t ] = y ( t − Δ t ) [ 1 + 1 2 λ Δ t ] y ( t ) y ( t − Δ t ) = 1 + 1 2 λ Δ t 1 − 1 2 λ Δ t = 1 + 1 2 λ 1 − 1 2 λ = 1 + 1 2 ( a + j b ) 1 − 1 2 ( a + j b ) \large{y(t) = y(t - \Delta t) + \frac{1}{2} \Big [ \dot{y}(t) + \dot{y}(t - \Delta t) \Big ] \, \Delta t \\ y(t) = y(t - \Delta t) + \frac{1}{2} \Big [ \lambda \, y(t) + \lambda \, y(t - \Delta t) \Big ] \, \Delta t \\ y(t) \Big[ 1 - \frac{1}{2} \lambda \, \Delta t \Big ] = y(t - \Delta t) \Big[ 1 + \frac{1}{2} \lambda \, \Delta t \Big] \\ \frac{y(t)}{y(t - \Delta t)} = \frac{ 1 + \frac{1}{2} \lambda \, \Delta t }{1 - \frac{1}{2} \lambda \, \Delta t } \\ = \frac{ 1 + \frac{1}{2} \lambda }{1 - \frac{1}{2} \lambda} = \frac{ 1 + \frac{1}{2} (a + jb) }{1 - \frac{1}{2} (a + jb)} }

Both the numerator and the denominator have imaginary parts of the same magnitude. By inspection, for a < 0 a < 0 , the denominator has a larger magnitude than the numerator, yielding convergence. For a > 0 a > 0 , the numerator has a larger magnitude than the denominator, yielding divergence. Thus, the convergence region contains the entire left-hand plane a < 0 a < 0 , and none of the right-hand plane a > 0 a > 0 . For a = 0 a = 0 , the numerator and denominator have the same magnitude, but different phase angles. This results in oscillation without growth or decay.

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