4.6.32 · D3Convolution theorem — proof, applications
Yeh note convolution theorem note ka hands-on companion hai. Wahan humne L { f ∗ g } = F ( s ) G ( s ) prove kiya tha. Yahan hum ise drill karenge jab tak har tarah ki problem — har sign, har degenerate input, har trap — ek baar mil na jaye.
Intuition "Convolution" ka matlab, ek aur baar, ek picture meinConvolution ( f ∗ g ) ( t ) = ∫ 0 t f ( τ ) g ( t − τ ) d τ blend hai: ek fixed present moment t g flipped copy ko f τ 0 t τ t − τ t
Figure s01 — picture padhna. Horizontal axis running clock τ t − τ ( τ , t − τ ) bold red diagonal woh hai jo t = 4 τ + ( t − τ ) = t τ t − τ t f ( τ ) g ( t − τ ) hi convolution hai.
Kisi bhi symbol se darne se pehle, yahan ek chota dictionary hai jo hum baar baar use karenge:
Definition Is page ke teen symbols
L { f } = F ( s ) Laplace transform — ek machine jo time-function f ( t ) s F ( s ) Laplace Transform — definition and properties mein bana hai.)L − 1 inverse — wahi machine ulti chalti hui, F ( s ) f ( t ) ∗ convolution , upar wala blend integral. Ise kabhi ordinary multiplication ×
Definition Standing hypothesis (causal / one-sided functions)Is page par sab kuch yeh assume karta hai ki f ( t ) g ( t ) causal hain: woh t < 0 0 t ≥ 0 L { f } = ∫ 0 ∞ e − s t f ( t ) d t t ≥ 0 Yeh limits ke liye kyun matter karta hai. Kyunki g ( u ) = 0 u < 0 g ( t − τ ) 0 τ > t ∫ 0 ∞ f ( τ ) g ( t − τ ) d τ ∫ 0 t t ∞ [ 0 , t ] L { f ∗ g } = F ( s ) G ( s )
Is topic ki har problem in cells mein se ek mein aati hai. Neeche ke worked examples mein se har ek us cell ke saath tagged hai jo woh cover karta hai, aur saath milke yeh sab ko hit karte hain.
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Cell (scenario class)
Isme kya tricky hai
Example
C1
Degenerate input : constant 1 Kya f ∗ 1 f
Ex 1
C2
Dono factors polynomial (1/ s m ⋅ 1/ s n Pure power algebra, koi trig nahi
Ex 2
C3
Repeated factor 1/ ( s 2 + a 2 ) 2 Same trig apne aap se blend ho raha hai
Ex 3
C4
Mixed exponential × trig Integral mein sign bookkeeping
Ex 4
C5
Do alag exponentials (e a t , e b t Degenerate limit a → b Ex 5
C6
ODE solve karo convolution ses Ex 6
C7
Volterra integral equation Hidden ∗
Ex 7
C8
Word problem / real system (impulse response)Physics → convolution translate karo
Ex 8
C9
Exam twist : commutativity + kaun sa factor flip karenAasaan integrand chuno
Ex 9
Jahan ek cell ka koi limiting / degenerate sub-case hai (jaise a → b
t 2 ∗ 1 Pehle forecast karo: andaza lagao — kya t 2 ∗ 1 t 2
Step 1. Pieces identify karo: f ( τ ) = τ 2 g ( t − τ ) = 1 1 Yeh step kyun? Convolution ko dono factors running variable τ 1
Step 2. Integral likho:
t 2 ∗ 1 = ∫ 0 t τ 2 ⋅ 1 d τ . Yeh step kyun? ( f ∗ g ) ( t ) = ∫ 0 t f ( τ ) g ( t − τ ) d τ g ≡ 1 1 literally 0 t hai — yahi intuition rakho.
Step 3. Integrate karo:
∫ 0 t τ 2 d τ = 3 t 3 .
Verify (theorem se): L { t 2 } = s 3 2 L { 1 } = s 1 s 4 2 L − 1 { s 4 2 } = 3 ! 2 t 3 = 3 t 3
Moral: 1 nahi hai — yeh integration hai. Tumhara "sirf t 2
Worked example Convolution seL − 1 { s 3 s 2 1 } = L − 1 { s 5 1 } Forecast: do power factors → answer ek single power of t
Step 1. Split karo: F ( s ) = s 3 1 ⇒ f ( t ) = 2 t 2 G ( s ) = s 2 1 ⇒ g ( t ) = t Yeh step kyun? L { t n } = s n + 1 n ! s 3 1 ↔ 2 ! t 2 s 2 1 ↔ 1 ! t 1
Step 2. Convolve karo:
( f ∗ g ) ( t ) = ∫ 0 t 2 τ 2 ( t − τ ) d τ . Yeh step kyun? Theorem apply karo; f ( τ ) = τ 2 /2 g g ( t − τ ) = t − τ
Step 3. Expand karo aur integrate karo:
2 1 ∫ 0 t ( t τ 2 − τ 3 ) d τ = 2 1 [ 3 t t 3 − 4 t 4 ] = 2 1 ⋅ 12 t 4 = 24 t 4 .
Verify: L { t 4 /24 } = 24 s 5 4 ! = 24 s 5 24 = s 5 1 4 1/ s 5
y ′′ + 4 y = 0 yeh chahiye: L − 1 { ( s 2 + 4 ) 2 1 } Forecast: sin squared denominator. sin t cos
Step 1. Pehchano ( s 2 + 4 ) 2 1 = s 2 + 4 1 ⋅ s 2 + 4 1 L − 1 { s 2 + 4 1 } = 2 sin 2 t Yeh step kyun? L { sin a t } = s 2 + a 2 a a = 2 s 2 + 4 2 2
Step 2. g g ( t ) = 2 1 sin 2 t y = ∫ 0 t 2 1 sin ( 2 τ ) ⋅ 2 1 sin ( 2 ( t − τ ) ) d τ = 4 1 ∫ 0 t sin 2 τ sin ( 2 t − 2 τ ) d τ . Yeh step kyun? Direct application; dono arguments 2 τ 2 ( t − τ ) 2 t
Step 3. Product-to-sum: sin A sin B = 2 1 [ cos ( A − B ) − cos ( A + B )] A = 2 τ B = 2 t − 2 τ A − B = 4 τ − 2 t A + B = 2 t y = 8 1 ∫ 0 t [ cos ( 4 τ − 2 t ) − cos 2 t ] d τ . Yeh step kyun? cos 2 t τ
Step 4. Integrate karo:
8 1 [ 4 s i n ( 4 τ − 2 t ) − τ cos 2 t ] 0 t = 8 1 [ 4 s i n 2 t − s i n ( − 2 t ) − t cos 2 t ] = 8 1 [ 4 2 s i n 2 t − t cos 2 t ] = 8 1 [ 2 s i n 2 t − t cos 2 t ] . Yeh step kyun? Upper limit τ = t sin ( 4 t − 2 t ) = sin 2 t τ = 0 sin ( − 2 t ) = − sin 2 t sin 2 t − ( − sin 2 t ) = 2 sin 2 t 2 sin 2 t /4 = sin 2 t /2
Step 5. 8 1 y = 8 1 ⋅ 2 s i n 2 t − 8 1 ⋅ t cos 2 t = 16 s i n 2 t − 8 t c o s 2 t . Yeh step kyun? Pehle term ke liye 8 1 ⋅ 2 1 = 16 1 8 1 ⋅ 1 = 8 1 alag hain. Phir 16 1 y = 16 sin 2 t − 8 t cos 2 t = 16 1 ( sin 2 t − 2 t cos 2 t )
Verify: jaana-maana formula L − 1 { ( s 2 + a 2 ) 2 1 } = 2 a 3 1 ( sin a t − a t cos a t ) a = 2 16 1 ( sin 2 t − 2 t cos 2 t )
L − 1 { ( s − 1 ) ( s 2 + 1 ) 1 } Forecast: pehle factor se e t sin t
Step 1. Split karo: F = s − 1 1 ⇒ f ( t ) = e t G = s 2 + 1 1 ⇒ g ( t ) = sin t Yeh step kyun? L { e a t } = s − a 1 a = 1 L { sin t } = s 2 + 1 1
Step 2. g ∫ 0 t e τ sin ( t − τ ) d τ . Yeh step kyun? Exponential ko f ( τ ) = e τ t − τ
Step 3. sin ( t − τ ) = sin t cos τ − cos t sin τ τ sin t ∫ 0 t e τ cos τ d τ − cos t ∫ 0 t e τ sin τ d τ . Yeh step kyun? sin t , cos t τ depend nahi karte , isliye integral ke liye yeh constants hain.
Step 4. Standard integrals use karo ∫ e τ cos τ d τ = 2 1 e τ ( cos τ + sin τ ) ∫ e τ sin τ d τ = 2 1 e τ ( sin τ − cos τ ) 0 t ∫ 0 t e τ cos τ d τ = 2 1 ( e t ( cos t + sin t ) − 1 ) , ∫ 0 t e τ sin τ d τ = 2 1 ( e t ( sin t − cos t ) + 1 ) .
Step 5. Substitute karo aur simplify karo (e t sin t cos t 2 1 e t − 2 1 cos t − 2 1 sin t
Independent partial-fraction inversion se verify karo (koi sign fudging nahi). Likho ( s − 1 ) ( s 2 + 1 ) 1 = s − 1 A + s 2 + 1 B s + C
A s = 1 A = s 2 + 1 1 s = 1 = 2 1 Denominators clear karo: 1 = A ( s 2 + 1 ) + ( B s + C ) ( s − 1 ) s 2 0 = A + B ⇒ B = − 2 1 s = 0 1 = A − C ⇒ C = A − 1 = − 2 1
Isliye ( s − 1 ) ( s 2 + 1 ) 1 = s − 1 1/2 + s 2 + 1 − 2 1 s − 2 1 L − 1 { 1/ ( s − 1 )} = e t L − 1 { s / ( s 2 + 1 )} = cos t L − 1 { 1/ ( s 2 + 1 )} = sin t 2 1 e t − 2 1 cos t − 2 1 sin t . cos t aur sin t − 2 1 e t badhta hai.)
a = b L − 1 { ( s − a ) ( s − b ) 1 } b → a Forecast: do alag decays e a t , e b t difference expect karo. Jab dono equal ho jaate hain tab kya hoga? Step 4 se pehle guess karo.
Step 1. f ( t ) = e a t g ( t ) = e b t Yeh step kyun? Har factor 1/ ( s − a ) 1/ ( s − b )
Step 2. Convolve karo:
∫ 0 t e a τ e b ( t − τ ) d τ = e b t ∫ 0 t e ( a − b ) τ d τ . Yeh step kyun? e b t τ ( a − b ) τ
Step 3. Integrate karo (valid kyunki a = b e b t ⋅ a − b e ( a − b ) t − 1 = a − b e a t − e b t . a − b e a t − e b t ( a = b )
Step 4 — degenerate case b → a Formula 0/0 b = a ∫ 0 t 1 d τ = t lim b → a a − b e a t − e b t = t e a t . Yeh step kyun? Jab a = b ( a − b ) τ = 0 1 t t e a t L − 1 { ( s − a ) 2 1 } = t e a t
Verify (general): L { a − b e a t − e b t } = a − b 1 ( s − a 1 − s − b 1 ) = ( s − a ) ( s − b ) 1 Verify (degenerate): L { t e a t } = ( s − a ) 2 1
y ′′ − y = e t y ( 0 ) = 0 , y ′ ( 0 ) = 0 Solving Linear ODEs with Laplace Transforms ).
Forecast: forcing e t e t resonant hai. Ek t e t
Step 1. Dono sides ka Laplace lo L { y ′′ } = s 2 Y − sy ( 0 ) − y ′ ( 0 ) s 2 Y − Y = s − 1 1 ⇒ ( s 2 − 1 ) Y = s − 1 1 . Yeh step kyun? Zero initial conditions boundary terms ko khatam kar dete hain, ODE ko algebra mein badal dete hain.
Step 2. Isolate karo aur factor karo: s 2 − 1 = ( s − 1 ) ( s + 1 ) Y = ( s − 1 ) 2 ( s + 1 ) 1 = F ( s − 1 ) 2 1 ⋅ G s + 1 1 . Yeh step kyun? Y product ke roop mein likhna ek badi partial fraction ki jagah convolution theorem ko invite karta hai.
Step 3. Inverses: f ( t ) = t e t g ( t ) = e − t
Step 4. Convolve karo:
y = ∫ 0 t τ e τ e − ( t − τ ) d τ = e − t ∫ 0 t τ e 2 τ d τ . Yeh step kyun? Exponentials combine karo: e τ e − ( t − τ ) = e − t e 2 τ
Step 5. Integration by parts se integrate karo ∫ 0 t τ e 2 τ d τ = [ 2 τ e 2 τ − 4 1 e 2 τ ] 0 t = 2 t e 2 t − 4 1 e 2 t + 4 1 y = e − t ( 2 t e 2 t − 4 1 e 2 t + 4 1 ) = 2 t e t − 4 1 e t + 4 1 e − t . y = 2 1 t e t − 4 1 e t + 4 1 e − t
Verify: y ′′ − y y = 2 1 t e t − 4 1 e t + 4 1 e − t y ′′ − y = e t y ( 0 ) = − 4 1 + 4 1 = 0 y ′ ( 0 ) = 0 t e t
y ( t ) = e − t + ∫ 0 t y ( τ ) ( t − τ ) d τ Volterra Integral Equations ).
Forecast: integral y ∗ t y t
Step 1. Hidden convolution pehchano: ∫ 0 t y ( τ ) ( t − τ ) d τ = ( y ∗ t ) ( t ) L { t } = s 2 1 Yeh step kyun? Kernel ( t − τ ) t − τ g ( t ) = t
Step 2. Poori equation ka Laplace lo (L { y } = Y L { e − t } = s + 1 1 Y = s + 1 1 + Y ⋅ s 2 1 . Yeh step kyun? Convolution → product integral equation ko ordinary algebra mein badal deta hai.
Step 3. Y Y ( 1 − s 2 1 ) = s + 1 1 ⇒ Y ⋅ s 2 s 2 − 1 = s + 1 1 ⇒ Y = ( s + 1 ) ( s 2 − 1 ) s 2 = ( s + 1 ) 2 ( s − 1 ) s 2 . Yeh step kyun? Saare Y s 2 − 1 = ( s − 1 ) ( s + 1 ) ( s + 1 ) 2
Step 4. Partial fractions: Y = s − 1 A + s + 1 B + ( s + 1 ) 2 C Yeh step kyun? Repeated factor ( s + 1 ) 2 B C
A s = 1 A = ( s + 1 ) 2 s 2 s = 1 = 4 1 C s = − 1 C = s − 1 s 2 s = − 1 = − 2 1 = − 2 1 B s = 0 ( 1 ) 2 ( − 1 ) 0 = 0 = − 1 A + 1 B + 1 C = − 4 1 + B − 2 1 ⇒ B = 4 3
Isliye A = 4 1 B = 4 3 C = − 2 1
Step 5. Term-by-term invert karo, L − 1 { 1/ ( s + 1 ) 2 } = t e − t Yeh step kyun? Har simple piece ka jaana-maana inverse hai; L − 1 { 1/ ( s − 1 )} = e t L − 1 { 1/ ( s + 1 )} = e − t t e − t y = 4 1 e t + 4 3 e − t − 2 1 t e − t . y = 4 1 e t + 4 3 e − t − 2 1 t e − t
Verify: t = 0 y ( 0 ) = 4 1 + 4 3 − 0 = 1 t = 0 e 0 + 0 = 1 y ( s + 1 ) 2 ( s − 1 ) s 2
Worked example Paani ek leaky tank meinconstant rate f ( t ) = 1 t = 0 τ e − ( t − τ ) t Forecast: shuruaat mein bharega; eventually inflow leak balance karega. Kis steady level ki umeed hai? Guess karo.
Step 1. Model: abhi ki quantity sabhi past instants τ y ( t ) = ∫ 0 t = 1 f ( τ ) survival e − ( t − τ ) d τ = ( 1 ∗ e − t ) ( t ) . Yeh step kyun? Yeh hi convolution hai — survival kernel e − ( t − τ ) impulse response hai (dekho Transfer Functions and Impulse Response ). Note karo e − ( t − τ ) = 0 t ≥ τ [ 0 , t ]
Step 2. Directly evaluate karo. τ e − t y ( t ) = ∫ 0 t e − ( t − τ ) d τ = e − t ∫ 0 t e τ d τ = e − t [ e τ ] 0 t = e − t ( e t − 1 ) = 1 − e − t . Yeh step kyun? e − ( t − τ ) = e − t e τ e τ τ y ( t ) = 1 − e − t
Step 3 — limiting behaviour (sabhi cases). t = 0 y = 1 − e 0 = 0 t → ∞ e − t → 0 y → 1 t > 0 0 1
Verify (theorem se): L { 1 } = s 1 L { e − t } = s + 1 1 s ( s + 1 ) 1 = s 1 − s + 1 1 1 − e − t Units: integrand hai (L/s)·(dimensionless survival)·(s) = L. ✓
t 3 ∗ cos t Forecast: polynomial flip karne par t 3 cos
Step 1. Commutativity kehti hai f ∗ g = g ∗ f t 3 ∗ cos t = ∫ 0 t τ 3 cos ( t − τ ) d τ = ∫ 0 t cos τ ( t − τ ) 3 d τ . Yeh step kyun? Parent proof se, u = t − τ
Step 2. Pehli form lo (cos ∫ 0 t τ 3 cos ( t − τ ) d τ L { t 3 } = s 4 6 , L { cos t } = s 2 + 1 s ⇒ L { t 3 ∗ cos t } = s 4 6 ⋅ s 2 + 1 s = s 3 ( s 2 + 1 ) 6 . Yeh step kyun? s t
Step 3. s 3 ( s 2 + 1 ) 6 s 3 ( s 2 + 1 ) 6 = s 3 6 − s 6 + s 2 + 1 6 s Yeh step kyun? s 3 ( s 2 + 1 ) 6/ s 3 − 6/ s 6 s / ( s 2 + 1 )
Step 4. Term-by-term invert karo:
t 3 ∗ cos t = 3 t 2 − 6 + 6 cos t . t 3 ∗ cos t = 3 t 2 − 6 + 6 cos t
Verify: t = 0 ∫ 0 0 ( ⋯ ) = 0 0 − 6 + 6 = 0 L { 3 t 2 − 6 + 6 cos t } = s 3 6 − s 6 + s 2 + 1 6 s = s 3 ( s 2 + 1 ) 6
Lesson (the twist): dono flip legal hain, lekin s
Recall Kaun sa cell kaun sa hai?1 e a t , e b t a − b e a t − e b t t e a t ∫ 0 t y ( τ ) k ( t − τ ) d τ Y ( s ) K ( s ) e − t 1
Mnemonic Smart route chuno"s t
Product F times G in s world
Convolution blend in t world
Convolve with 1 equals integrate
Two exponentials difference quotient
Degenerate limit gives t times e
ODE forcing solved by blend
Volterra equation becomes algebra
Impulse response of a real system