1. **Problem Statement:** Determine if each system is time-invariant or time-variant. 2. **Key Concept:** A system is time-invariant if a time shift in the input causes the same time shift in the output. Mathematically, if $y(n) = T\{x(n)\}$, then for any shift $k$, $T\{x(n-k)\} = y(n-k)$ must hold. 3. **System (i):** $y(n) = x(-n)$ - Input shift: $x(n-k)$ - Output: $y_k(n) = x(-(n-k)) = x(-n + k)$ - Compare with shifted output: $y(n-k) = x(-(n-k)) = x(-n + k)$ - Since $y_k(n) = y(n-k)$, system is **time-invariant**. 4. **System (ii):** $y(n) = x(n) - x(n-1)$ - Input shift: $x(n-k)$ - Output: $y_k(n) = x(n-k) - x(n-k-1)$ - Shifted output: $y(n-k) = x(n-k) - x(n-k-1)$ - Since $y_k(n) = y(n-k)$, system is **time-invariant**. 5. **System (iii):** $y(n) = n x(n)$ - Input shift: $x(n-k)$ - Output: $y_k(n) = n x(n-k)$ - Shifted output: $y(n-k) = (n-k) x(n-k)$ - Since $y_k(n) \neq y(n-k)$ (due to factor $n$ vs $n-k$), system is **time-variant**. 6. **System (iv):** $y(t) = \sin[x(t)]$ - Input shift: $x(t-\tau)$ - Output: $y_\tau(t) = \sin[x(t-\tau)]$ - Shifted output: $y(t-\tau) = \sin[x(t-\tau)]$ - Since $y_\tau(t) = y(t-\tau)$, system is **time-invariant**. 7. **System (v):** $y(t) = t \cdot x(t)$ - Input shift: $x(t-\tau)$ - Output: $y_\tau(t) = t \cdot x(t-\tau)$ - Shifted output: $y(t-\tau) = (t-\tau) \cdot x(t-\tau)$ - Since $y_\tau(t) \neq y(t-\tau)$ (due to factor $t$ vs $t-\tau$), system is **time-variant**. 8. **System (vi):** $y(t) = x(t) \cos(200 \pi t)$ - Input shift: $x(t-\tau)$ - Output: $y_\tau(t) = x(t-\tau) \cos(200 \pi t)$ - Shifted output: $y(t-\tau) = x(t-\tau) \cos[200 \pi (t-\tau)]$ - Since $\cos(200 \pi t) \neq \cos[200 \pi (t-\tau)]$, system is **time-variant**. **Final answers:** (i) Time-invariant (ii) Time-invariant (iii) Time-variant (iv) Time-invariant (v) Time-variant (vi) Time-variant