Math Editor

Describing Function

Prepared by: Ibrahim Mohamed El-bastawisi

A describing function describes the behavior of a nonlinear element for purely sinusoidal excitation.

N(A,ω)=x(ωt)y(ωt)​ where x is a sinusoidal input of amplitude A : x(ωt)=Asinωt

if we expand the output in a Fourier Series and consider only the first harmonic, the describing function becomes:

N(A,ω)=x(ωt)y(ωt)​=Asinωtacosωt+bsinωt​=A∠0a∠90+b∠0​=A√a2+b2​​∠tan−1ba​

On-Off

x(ωt)=Asinωt y(x)={M−M​x>0x<0​ y(ωt)={M−M​Asinωt>0Asinωt<0​

Find Fourier series fundamental component:

y(ωt)={M−M​Asinωt>0Asinωt<0​

y(ωt)={M−M​0<ωt<ππ<ωt<2π​

a=0

b=π4M​∫0π/2​sinωtdωt=π4M​

N(A)=πA4M​

−N(A)1​=−4MπA​

On-Off with Dead Zone

x(ωt)=Asinωt y(x)=⎩⎪⎨⎪⎧​M−M0​x>δx<−δotherwise​ y(ωt)=⎩⎪⎨⎪⎧​M−M0​Asinωt>δAsinωt<−δotherwise​

Find Fourier series fundamental component:

y(ωt)=⎩⎪⎨⎪⎧​M−M0​Asinωt>δAsinωt<−δotherwise​

let γ∷x(γ)=δ

y(ωt)=⎩⎪⎨⎪⎧​M−M0​γ<ωt<π−γπ+γ<ωt<2π−γotherwise​

a=0

b=π4M​∫γπ/2​sinωtdωt=π4M​cosγ

but x(γ)=δ⟹Asinγ=δ

sinγ=Aδ​

cosγ=A√A2−δ2​​=√1−(Aδ​)2​

γ=sin−1Aδ​

N(A)=πA4M​√1−(Aδ​)2​

−N(A)1​=−4M√1−(δ/A)2​πA​

On-Off with Hysteresis

x(ωt)=Asinωt y(x)=⎩⎪⎨⎪⎧​M−M??​x>δx<−δotherwise​ y(ωt)=⎩⎪⎨⎪⎧​M−M??​Asinωt>δAsinωt<−δotherwise​

Find Fourier series fundamental component:

let γ∷x(γ)=δ

y(ωt)={M−M​γ<ωt<π+γπ+γ<ωt<2π+γ​

a=π2​∫γπ+γ​Mcosωtdωt=π2M​(sin(π+γ)−sinγ)=π−4M​sinγ

b=π2​∫γπ+γ​Msinωtdωt=π2M​(−cos(π+γ)+cosγ)=π4M​cosγ

N(A)=A√a2+b2​​∠tan−1ba​

√a2+b2​=√(π−4M​sinγ)2+(π4M​cosγ)2​=π4M​

tan−1ba​=tan−1(4M/π)cosγ−(4M/π)sinγ​=−γ

N(A)=πA4M​∠−γ

but x(γ)=δ⟹Asinγ=δ

sinγ=Aδ​

cosγ=A√A2−δ2​​=√1−(Aδ​)2​

γ=sin−1Aδ​

N(A)=πA4M​∠−sin−1Aδ​

N(A)−1​=−4MπA​∠γ=−4MπA​(cosγ+jsinγ)

N(A)−1​=−4MπA​(√1−(Aδ​)2​+jAδ​)

Saturation

x(ωt)=Asinωt y(x)=⎩⎪⎨⎪⎧​M−MKx​x>δx<−δotherwise​ y(ωt)=⎩⎪⎨⎪⎧​M−MKAsinωt​Asinωt>δAsinωt<−δotherwise​

Find Fourier series fundamental component:

let γ∷x(γ)=δ

y(ωt)=⎩⎪⎨⎪⎧​M−MKAsinωt​γ<ωt<π−γπ+γ<ωt<2π−γotherwise​

a=0

b=π4​(∫0γ​KAsin2ωt+∫γπ/2​Msinωtdωt)

=π4​(∫0γ​2KA​(1−cos2ωt)+∫γπ/2​Msinωtdωt)

=π4​(2KA​[ωt−2sin2ωt​]0γ​+[−Mcosωt]γπ/2​)

=π4​(2KA​(γ−2sin2γ​)+Mcosγ)

but M=Kδ

b=π4K​(2A​(γ−2sin2γ​)+δcosγ)

and x(γ)=δ⟹Asinγ=δ

b=π4KA​(21​(γ−2sin2γ​)+sinγcosγ)

and sin2γ=2sinγcosγ

b=π4KA​(21​(γ−sinγcosγ)+sinγcosγ)

b=π2KA​(γ+sinγcosγ)

sinγ=Aδ​

cosγ=A√A2−δ2​​=√1−(Aδ​)2​

γ=sin−1Aδ​

b=π2KA​(sin−1Aδ​+Aδ​√1−(Aδ​)2​)

N(A)=π2K​(sin−1Aδ​+Aδ​√1−(Aδ​)2​)

let Aδ​=x⟹KN​(x)=π2​(sin−1x+x√1−x2​)

Dead Zone

x(ωt)=Asinωt y(x)=⎩⎪⎨⎪⎧​K(x−δ)K(x+δ)0​x>δx<−δotherwise​ y(x)=⎩⎪⎨⎪⎧​K(Asinωt−δ)K(Asinωt+δ)0​x>δx<−δotherwise​

Find Fourier series fundamental component:

let γ∷x(γ)=δ

y(ωt)=⎩⎪⎨⎪⎧​K(Asinωt−δ)K(Asinωt+δ)0​γ<ωt<π−γπ+γ<ωt<2π−γotherwise​

a=0

b=π4​∫γπ/2​K(Asinωt−δ)sinωt

but x(γ)=δ⟹Asinγ=δ

b=π4AK​∫γπ/2​sin2ωt−sinγsinωt

=π4AK​∫γπ/2​21​−21​cos2ωt−sinγsinωt

=π4AK​[2ωt​−41​sin2ωt+sinγcosωt]γπ/2​

b=π4AK​[(4π​−0+0)−(2γ​−41​sin2γ+sinγcosγ)]

but sin2γ=2sinγcosγ

b=π4AK​(4π​−2γ​−21​sinγcosγ)

sinγ=Aδ​

cosγ=A√A2−δ2​​=√1−(Aδ​)2​

γ=sin−1Aδ​

b=π4AK​(4π​−21​(sin−1Aδ​+Aδ​√1−(Aδ​)2​))

N(A)=π4K​(4π​−21​(sin−1Aδ​+Aδ​√1−(Aδ​)2​))

let Aδ​=x⟹KN​(x)=π4​(4π​−21​(sin−1x+x√1−x2​))